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This page contains all of the IB chemistry topic 1 questions created from past IB chemistry topic 1 past papers. IB chem topic 1 covers the IB chemistry moles content from the IB chemistry course. The sub-topics included are shown below, covering the IB chemistry topic 1 areas of: atoms, mixtures, atomic mass, molecular mass, empirical formula, limiting reactants, Avogadro's law, and standard solutions.

Our IB chem topic 1 questions on IB chemistry stoichiometry test your topic 1 syllabus knowledge required for the IB chemistry topic 1 questions in the exam. They will also prepare you well for IB chemistry topic 1 past paper questions!

When the equation below is balanced, the sum of all the coefficients is

(CH_{3})_{2}NNH_{2} + N_{2}O_{4} → N_{2} + CO_{2} + H_{2}O

Select an answer from the options

Correct, the answer is C!

The equation balances to: (CH_{3})_{2}NNH_{2} + 2N_{2}O_{4} → 3N_{2} + 2CO_{2} + 4H_{2}O

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Report ErrorFor the following equation, the sum of all the balanced coefficients is

CH_{3}CH_{2}COCH_{3}_{(l)} + O_{2} _{(g)} → CO_{2} _{(g)} + H_{2}O_{(l)}

Select an answer from the options

Correct, the answer is D!

The equation balances to: 2CH_{3}CH_{2}COCH_{3}_{(l)} + 11O_{2} _{(g)} → 8CO_{2} _{(g)} + 8H_{2}O_{(l)}

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Report ErrorWhat is the molar ratio of O_{2} to H_{2}O when the equation below is balanced?

NH_{3}_{(g)} + O_{2} _{(g)} → NO_{(g)} + H_{2}O_{(g)}

Select an answer from the options

Correct, the answer is B!

The equation balances to: 4NH_{3}_{(g)} + 5O_{2} _{(g)} → 4NO_{(g)} + 6H_{2}O_{(g)}

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Report ErrorWrite a balanced equation for the complete combustion of hexane, C_{6}H_{14}. The sum of the coefficients in this balanced equation would be

Select an answer from the options

Correct, the answer is D!

The equation would be: 2C_{6}H_{14} + 19O_{2} → 12CO_{2} + 14H_{2}O.

**Note:** A useful tip is to balance C, H and O in that order. If you need to, use a decimal to make the O balance, then double everything after!

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Report ErrorFor the below equation, what is the value of *b* when *a* = 4?

*a*C_{2}H_{3}Cl_{(g)} + *b*O_{2} → *c*CO_{2} _{(g)} + *d*H_{2}O_{(g)} + *e*HCl_{(g)}

Select an answer from the options

Correct, the answer is A!

The equation would be: 4C_{2}H_{3}Cl_{(g)} + 10O_{2} → 8CO_{2} _{(g)} + 4H_{2}O_{(g)} + 4HCl_{(g)}

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Report ErrorIf the following reaction produced 3.15 mol of oxygen, how many moles of KCl would be produced?

2KClO_{3} → 2KCl + 3O_{2}

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Correct, the answer is A!

O_{2} and KCl are in a 3:2 ratio. So, moles of KCl is 3.15 ÷ 3 * 2 = 2.1

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Report ErrorCalculate the mass of H_{2} when 25 g of Al reacts with excess HCl as per the equation shown

2Al_{(s)} + 6HCl_{(aq)} → 2AlCl_{3}_{(aq)} + 3H_{2}_{(g)}

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_{r}. So, moles of Al is 25 ÷ 27 = 0.926

Al and H_{2} are in a 2:3 ratio, so moles of H_{2} is 0.926 ÷ 2 * 3 = 1.38

Mass = Moles * M_{r}. So, mass of H_{2} is therefore 1.38 * 2 = 2.77

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Report ErrorWrite a balanced equation for the complete combustion of pentane, then calculate the mass of oxygen required to burn 14.4g of C_{5}H_{12}

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Correct, the answer is C!

The equation would be: C_{5}H_{12} + 8O_{2} → 5CO_{2} + 6H_{2}O

Moles of C_{5}H_{12} is 14.4 ÷ 72 = 0.2

C_{5}H_{12} is in a 1:8 ratio with O_{2} , so moles of O_{2} is 0.2 * 8 = 1.6

Mass = Moles * M_{r}. So, mass of O_{2} is therefore 1.6 * 32 = 51.2

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Report ErrorFor the reaction: C_{5}H_{12} + 8O_{2} → 5CO_{2} + 6H_{2}O

How many moles of water would form from 28.8 g of pentane?

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Correct, the answer is B!

Moles = Mass ÷ M_{r}. So, moles of pentane is 28.8 ÷ 72 = 0.4

Pentane and water are in a 1:6 ratio, so moles of water is 0.4 * 6 = 2.4

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Report ErrorWhat is the minimum mass of O_{2} required for combustion of 3.2 grams of methane according to the equation shown below?

CH_{4} + 2O_{2} → CO_{2} + 2H_{2}O

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_{r}. So, moles of methane is 3.2 ÷ 16 = 0.2

Methane and O_{2} are in a 1:2 ratio, so moles of O_{2} is 0.2 * 2 = 0.4

Mass = Moles * M_{r}. So, mass of O_{2} is therefore 0.4 * 32 = 12.8

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Report ErrorUsing the equation for the oxidation of NH_{3} shown below, calculate the water produced (grams) from oxidising 3.4g of NH_{3}

4NH_{3} + 5O_{2} → 4NO + 6H_{2}O

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_{r}. So, moles of NH_{3} is 3.4 ÷ 17 = 0.2

NH_{3} and water are in a 4:6 ratio, so moles of water is 0.2 ÷ 4 * 6 = 0.3

Mass = Moles * M_{r}. So, mass of water is therefore 0.3 * 18 = 5.4

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Report ErrorAl reacts with HCl to form hydrogen gas as per the equation below:

2Al_{(s)} + 6HCl_{(aq)} → 3H_{2}_{(g)} + 2AlCl_{3}_{(aq)}

Which expression calculates the number of moles of H_{2} formed from 0.12 moles of Al? Assume the
HCl is in excess

Select an answer from the options

Correct, the answer is A!

Al and H_{2} are in a 2:3 ratio, so moles of H_{2} is (0.12 ÷ 2) * 3. This is the same as 0.12 * (3 ÷ 2)

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Report ErrorFor the reaction shown, how much MnO_{2} is required to produce 50 g of KMnO_{4} ?

2MnO_{2} + 4KOH + O_{2} + Cl_{2} → 2KMnO_{4} + 2KCl + 2H_{2}O

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_{r}. So, moles of KMnO_{4} is 50 ÷ 158 = 0.316

KMnO_{4} and MnO_{2} are in a 2:2 ratio, so, moles are identical

Mass = Moles * M_{r}. So, mass of MnO_{2} is therefore 0.316 * 87 = 27.5

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Report ErrorFor the balanced equation: 3BaCl_{2}_{(aq)} + 2Na_{3} PO_{4} _{(aq)} → Ba_{3}(PO_{4})_{2}_{(s)} + 6NaCl_{(aq)}

How many moles of sodium chloride could be collected using 20 moles of BaCl_{2} in excess Na_{3} PO_{4} ?

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Correct, the answer is D!

BaCl_{2} and sodium chloride are in a 3:6 ratio, so moles of sodium chloride is 20 ÷ 3 * 6 = 40

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Report ErrorUsing the equation: 2Al_{(s)} + 3Cl_{2}_{(g)} → Al_{2}Cl_{6}_{(s)}

Calculate the mass of Al required to produce 34.2 g of Al_{2}Cl_{6}

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Moles = Mass ÷ M_{r}. So, moles of Al_{2}Cl_{6} is 34.2 ÷ 267 = 0.128 ✓

Al_{2}Cl_{6} and Al are in a 1:2 ratio, so moles of Al is 0.128 * 2 = 0.256 ✓

Mass = Moles * M_{r}. So, mass of Al is therefore 0.256 * 27 = 6.917g → 6.92g (3sf) ✓

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Report ErrorCalculate the mass of Fe_{2}O_{3} required to produce 3632 kg of Fe as per the equation shown below:

Fe_{2}O_{3} + 3CO → 2Fe + 3CO_{2}

Write out your answer in the box

Moles = Mass ÷ M_{r}. So, moles of Fe is 3632 ÷ 56 = 64.857. ✓

Fe and Fe_{2}O_{3} are in a 2:1 ratio, so moles of Fe_{2}O_{3} is 64.857 ÷ 2 = 32.428. ✓

Mass = Moles * M_{r}. So, mass of Fe_{2}O_{3} is therefore 32.428 * 160 = 5188.57kg → 5190kg (3sf) ✓

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Report ErrorWhich of these options has the greatest mass?

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Correct, the answer is D!

Mass = M_{r} * Moles. Substituting the relevant values from each option into this formula reveals that the last option has the greatest mass.

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Report ErrorThe number of moles in 250 g of water is:

Select an answer from the options

Correct, the answer is A!

Moles = Mass ÷ M_{r}. So, 250 ÷ 18 = 13.88

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Report ErrorThe mass of two molecules of water in grams would be:

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Correct, the answer is A!

Mass = M_{r} * Mole. So, 1 mole of water weighs 18g

1 mole of anything contains 6.02x10^{23} molecules, so 1 molecule weighs 18 ÷ 6.02x10^{23} = 2.99x10^{-23}

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Report ErrorHow many molecules are present in 18 g of H_{2}O?

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Correct, the answer is C!

1 mole of water weighs 18g. One mole of anything contains 6.02x10^{23} molecules

**Note:** 1 Mole of anything weighs the same as its M_{r}!

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Report ErrorWhat is the number of atoms in 0.20 mol of propyne, C_{3}H_{4}?

Select an answer from the options

Correct, the answer is C!

The number of molecules present is 0.2 * 6.02x10^{23} = 1.2x10^{23}

There are 7 atoms in each molecule so 7 * 1.2x10^{23} = 8.42x10^{23} atoms

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Report ErrorA single atom of an element has a mass of 1.06x10^{-22} grams. The element is:

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Correct, the answer is A!

The total weight of 1 mole of atoms would be 1.06x10^{-22} * 6.02x10^{-23} = 63.8

Since 1 mole of anything = Mr, this would be the Mr

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Report ErrorWhat is the mass of one molecule of propanol in grams?

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Correct, the answer is B!

1 mole weighs the same as Mr, which = 60

There are 6.02x10^{23} molecules in this, so 60 ÷ 6.02x10^{23} = 1x10^{-22}

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Report ErrorThe mass spectrum of Mg produced the data shown below. Using this information, the A_{r} of Mg would be

Mass/Charge | % Abundance |
---|---|

79 | 78.6 |

80 | 10.11 |

81 | 11.29 |

Select an answer from the options

Correct, the answer is B!

The formula to use here is (M_{r} * %) + (M_{r} * %) + (M_{r} * %)100 = 79.32

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Report ErrorWhich contains the largest number of molecules?

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Correct, the answer is B!

Whichever has the most moles will have the most molecules. Moles = Mass ÷ M_{r}. They all have the same mass, so whichever has the lowest M_{r} will be correct

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Report ErrorThe mass in grams of one molecule of C_{3}H_{7}Br is

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Correct, the answer is B!

1 mole weighs the same as Mr, which = 123

There are 6.02x10^{23} molecules in this, so 123 ÷ 6.02x10^{23} = 2.04x10^{-22}

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Report ErrorHalf a mole of carbon dioxide molecules would contain

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Correct, the answer is D!

1 mole of anything contains 6.02x10^{23} molecules, so 0.5 mole contains 0.5 * 6.02x10^{23} = 3.01x10^{23}

There are 3 atoms in each molecule, so 3.01x10^{23} * 3 = 9.03x10^{23}

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Report ErrorIn 0.250 moles of CH_{3}CHClOH, there are

Select an answer from the options

Correct, the answer is C!

1 mole of anything contains 6.02x10^{23} molecules, so 0.250 mole contains 0.250 * 6.02x10^{23} = 1.505x10^{23} molecules

There are 9 atoms in each molecule, so 1.505x10^{23} * 9 = 1.35x10^{24}

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Report ErrorWhich of the answers contains the smallest number of molecules?

Select an answer from the options

Correct, the answer is B!

Whichever has the fewest moles will have the fewest molecules. Moles = Mass ÷ M_{r}. They all have the same mass, so whichever has the highest M_{r} will be correct

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Report ErrorA sample of 6.44 g of O_{3} contains the same number of atoms as

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Correct, the answer is A!

The number of moles is Mass ÷ Mr, so 6.44 ÷ 48 = 0.134

The number of molecules is 0.134 * 6.02x10^{23} = 8.06x10^{22}

So, the number of atoms is 3 * 8.06x10^{22} = 2.42x10^{23}. Repeat this for each option

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Report ErrorThe mass in grams of one molecule of C_{10}H_{14}N_{2} (Nicotine) would be

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Correct, the answer is B!

1 mole weighs the same as Mr, which = 162.26

There are 6.02x10^{23} molecules in this, so 162.26 ÷ 6.02x10^{23} = 2.70x10^{-22}

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Report ErrorOne mole of CO_{2} molecules contains roughly

Select an answer from the options

Correct, the answer is D!

1 mole of anything contains 6.02x10^{23} molecules

There are 3 atoms in each molecule, so 6.02x10^{23} * 3 = 1.806x10^{24} atoms in 1 mole

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Report ErrorThe option which would contain 1.0x10^{23} atoms is

Select an answer from the options

Correct, the answer is B!

Mass ÷ M_{r} = Moles

Moles * 6.02x10^{23} = molecules

Molecules * number of atoms in a molecule = total atoms in that many moles. Repeat this for each

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Report ErrorThe mass in grams of two molecules of CO_{2} is

Select an answer from the options

Correct, the answer is C!

1 mole weighs the same as Mr, which = 44

There are 6.02x10^{23} molecules in this, so 44 ÷ 6.02x10^{23} = 7.308x10^{-23} is the mass of one molecule

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Report ErrorWhich of the following has the greatest mass?

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Correct, the answer is A!

8 moles of oxygen have a mass of 8 * 32 = 256. 1 mole of gallium has a mass of 1 * 70 = 70

Since it is diatomic, the number of moles of helium is 8x10^{25} * 2 ÷ 6.02x10^{23} = 265.78

So, the mass is 4 * 265.78 = 1063.12. Repeat for iron

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Report ErrorWhich of the answers would contain the fewest atoms?

Select an answer from the options

Correct, the answer is C!

For A, the number of moles is 1.5 ÷ 8 = 0.1875

The number of molecules is thus 0.1875 * 6.02x10^{23} = 1.13x10^{23}

There are 2 atoms in each molecule so 2 * 1.13x10^{23} = 2.26x10^{23} atoms. Repeat for the rest

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Report ErrorWhich of the following would contain the largest number of atoms?

Select an answer from the options

Correct, the answer is A!

For A, the number of moles is 1.5 ÷ 20 = 0.075

The number of molecules is thus 0.075 * 6.02x10^{23} = 4.515x10^{22}

There is 1 atom in each molecule so 4.515x10^{22} atoms. Repeat for the rest

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Report ErrorA sample of X with atomic mass 139.446 is made up of 60.4% of X-138 and 39.6% of X-142. If the mass of X-138 is 138.058, the mass of X-142 would be

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Correct, the answer is D!

The formula to use here is (M_{r} * %) + (M_{r} * %)100 = 139.446

So ((138.058 * 60.4) + (M_{r} * 39.6)) ÷ 100 = 139.446. Then rearrange

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Report ErrorWhich sample contains the smallest amount of oxygen?

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Correct, the answer is A!

Comparing the number of atoms would be easiest

So, moles * 6.02x10^{23} = molecules

Molecules * number of O atoms in each molecule = total number of O atoms

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Report ErrorHow many moles of C_{2}H_{6} are needed to obtain 6.0x10^{23} hydrogen atoms?

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Correct, the answer is A!

Number of molecules = 6.0x10^{22} ÷ 6 = 1x10^{22}

Number of moles = 1x10^{22} ÷ 6.02x10^{23} = 0.167

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Report ErrorHow many moles of O_{3} would contain 3.6x10^{22} molecules?

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Correct, the answer is B!

Moles = 3.6x10^{22} ÷ 6.02x10^{23} = 0.0598

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Report ErrorWhich of the answers is both an empirical and molecular formula?

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Correct, the answer is A!

Empirical formula cannot be divided, so B, C and D are wrong

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Report ErrorWhich of the following would have the greatest empirical formula mass?

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Correct, the answer is B!

A would be CH, B would be C_{2}H_{5}, C would be CH_{2}, and D would be CH_{3}

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Report ErrorA compound containing only C, H, O has the percentage by mass of 60 %, 8 %, 32 % respectively. The empirical formula would be?

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Correct, the answer is A!

Empirical formula are best calculated in a table:

C | H | O | |
---|---|---|---|

Mass/Percentage | 60 | 8 | 32 |

/\ ÷ M_{r} (Moles) |
5 | 8 | 2 |

/\ ÷ Smallest (Ratio) | 2.5 | 4 | 1 |

Final ratio | 5 | 8 | 2 |

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Report ErrorA compound has the empirical formula CH_{2} and an M_{r} in the region of 120 to 160. What is its most likely molecular formula?

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Correct, the answer is B!

The empirical formula has M_{r} of 14. 120 ÷ 14 = 8.6 and 160 ÷ 14 = 11.4. So, the M_{r} is roughly 9-11 times larger

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Report ErrorA compound consists of 87.5% N_{2} and 12.5% H_{2}. The molecular mass is 32.0 g mol^{-1}. What would be the molecular formula?

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Correct, the answer is D!

First work out the empirical formula, which is NH_{2}, as below:

N | H | |
---|---|---|

Mass/Percentage | 87.5 | 12.5 |

/\ ÷ M_{r} (Moles) |
6.25 | 12.5 |

/\ ÷ Smallest (Ratio) | 1 | 2 |

Final ratio | 1 | 2 |

The empirical formula has an empirical mass of 16. 32 ÷ 16 = 2

The answer is therefore twice the empirical formula

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Report ErrorThe molecular formula of a compound containing 85.7 % C and 14.3 % H_{2}, by mass, could be?

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Correct, the answer is B!

Work out the empirical formula, which is CH_{2}, as below:

C | H | |
---|---|---|

Mass/Percentage | 85.7 | 14.3 |

/\ ÷ M_{r} (Moles) |
7.14 | 14.3 |

/\ ÷ Smallest (Ratio) | 1 | 2 |

Final ratio | 1 | 2 |

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Report ErrorA compound contains 24 % Mg, 28 % Si and 48 % O_{2} . What is its empirical formula?

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Correct, the answer is D!

The empirical formula is MgSiO_{3} , as below:

Element | Mg | Si | O |
---|---|---|---|

Mass/Percentage | 24 | 28 | 48 |

/\ ÷ M_{r} (Moles) |
1 | 1 | 3 |

/\ ÷ Smallest (Ratio) | 1 | 1 | 3 |

Final ratio | 1 | 1 | 3 |

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Report ErrorWhat is the empirical formula for a compound with the molecular formula C_{12}H_{6}Cl_{6}?

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Correct, the answer is C!

All of the elements can be divided by 6, giving 2, 1, 1 respectively

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Report ErrorWhich answer shows the % by mass of C in C_{7}H_{5}(NO_{2})_{3}?

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Correct, the answer is B!

The total M_{r} is 227. C accounts for 84 of this. So as a %, (84 ÷ 227) * 100 = 37%

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Report ErrorThallium (0.203 g) is reacted completely with bromine gas to form 0.805 g of a compound containing only these two elements. What is its formula?

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Correct, the answer is C!

The mass of Br in the compound is 0.805 - 0.203 = 0.602. Calculate the empirical formula:

Element | Tl | Br |
---|---|---|

Mass/Percentage | 0.203 | 0.602 |

/\ ÷ M_{r} (Moles) |
9.93x10^{-4} |
7.52x10^{-3} |

/\ ÷ Smallest (Ratio) | 1 | 7.56 |

Final ratio | 1 | 7.5 |

**Note:** You can often round ratios from <1.3 down to 1, and >1.8 up to 2. However, keep ratios around .33, .5 and .75 and triple, double and quadruple the final ratio respectively

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Report ErrorA iodide of tungsten contains 55.1% iodine by mass. What is its simplest formula?

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Correct, the answer is D!

The % of tungsten in the compound is 100 - 55.1 = 44.9. Calculate the empirical formula:

Element | W | I |
---|---|---|

Mass/Percentage | 44.9 | 55.1 |

/\ ÷ M_{r} (Moles) |
0.244 | 0.434 |

/\ ÷ Smallest (Ratio) | 1 | 1.77 |

Final ratio | 4 | 7 |

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Report Error8 g of oxygen combine with a metal (A) of A_{r} 40.0 to give 18.0 g of product. What would be the empirical formula of the oxide formed?

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Correct, the answer is B!

The mass of A in the compound is 18 - 8 = 10. Calculate the empirical formula:

Element | A | O |
---|---|---|

Mass/Percentage | 10 | 8 |

/\ ÷ M_{r} (Moles) |
0.25 | 0.5 |

/\ ÷ Smallest (Ratio) | 1 | 2 |

Final ratio | 1 | 2 |

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Report ErrorA compound has an M_{r} of 112 g mol^{-1}. The empirical formula could never be

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Correct, the answer is D!

The empirical formula must have an empirical mass that is a factor of 112 (can divide into it exactly)

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Report ErrorA sample contains 0.500 g of H_{2}, 3.000 g of C and 17.75 g of Cl_{2}. What is the empirical formula?

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Correct, the answer is B!

The mass of A in the compound is 18 - 8 = 10. Calculate the empirical formula:

Element | C | Cl | H |
---|---|---|---|

Mass/Percentage | 3 | 17.75 | 0.5 |

/\ ÷ M_{r} (Moles) |
0.25 | 0.5 | 0.5 |

/\ ÷ Smallest (Ratio) | 1 | 2 | 2 |

Final ratio | 1 | 2 | 2 |

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Report ErrorA 20.0 g sample copper oxide, when heated in hydrogen, forms 2.52 g of water. The % mass of Cu in copper oxide would be

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Correct, the answer is D!

The moles of water would be 2.52 ÷ 18 = 0.14. This is the moles of oxygen in this water, and thus in the copper oxide (as it was not heated in air). The mass of oxygen would be 0.14 * 16 = 2.24. The mass of copper is therefore 20 - 2.24 = 17.76. This as a % of 20 is 88.8%

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Report ErrorWhen 11.864 g of lead oxide was heated with H_{2}, it was completely reduced to 10.758 g Pb. What is the empirical formula for the original lead oxide?

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Correct, the answer is D!

The mass of oxygen in the lead oxide is 11.864 – 10.758 = 1.106. Calculate the empirical formula:

Element | Pb | O |
---|---|---|

Mass/Percentage | 10.758 | 1.106 |

/\ ÷ M_{r} (Moles) |
0.0519 | 0.069 |

/\ ÷ Smallest (Ratio) | 1 | 1.33 |

Final ratio | 3 | 4 |

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Report ErrorDetermine the empirical formula of terbium peroxide given 0.140 mol of Tb and 0.245 mol of O_{2} in a sample

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Correct, the answer is D!

Calculate the empirical formula:

Element | Tb | O |
---|---|---|

Mass/Percentage | - | - |

/\ ÷ M_{r} (Moles) |
0.140 | 0.245 |

/\ ÷ Smallest (Ratio) | 1 | 1.75 |

Final ratio | 4 | 7 |

**Note:** You can often round ratios from <1.3 down to 1, and >1.8 up to 2. However, keep ratios around .33, .5 and .75 and triple, double and quadruple the final ratio respectively

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Report ErrorA hydrated salt contained 45.7% water, but the anhydrous salt contains: Na 29.8%; C 7.8%; O 73.4%. The formula of the hydrated salt would be

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Correct, the answer is D!

Calculate the empirical formula of the anhydrous salt, which is Na_{2}CO_{7} , as below:

Element | Na | C | O |
---|---|---|---|

Mass/Percentage | 29.8 | 7.8 | 73.4 |

/\ ÷ M_{r} (Moles) |
1.295 | 0.65 | 4.64 |

/\ ÷ Smallest (Ratio) | 1.99 | 1 | 7.14 |

Final ratio | 2 | 1 | 7 |

If the hydrated salt is 45.7% water, the Na_{2}CO_{7} (M_{r} 170) accounts for 54.3%

Thus, the water must have a total M_{r} of (170 ÷ 54.3) * 45.7 = 143.07

The number of water molecules that would make this M_{r} is 143.07 ÷ 18 = 7.95

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Report ErrorCitrazinic acid contains 46.46% C, 3.25% H_{2}, 9.03% N_{2} and 41.26% O_{2} . If there is only one N atom in the molecule, the total number of atoms combined would be?

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Correct, the answer is B!

Calculate the empirical formula of the acid, which is C_{6}H_{5}NO_{4} , as below:

Element | C | H | N | O |
---|---|---|---|---|

Mass/Percentage | 46.46 | 3.25 | 9.03 | 41.26 |

/\ ÷ M_{r} (Moles) |
3.87 | 3.25 | 0.645 | 2.578 |

/\ ÷ Smallest (Ratio) | 6 | 5.03 | 1 | 3.99 |

Final ratio | 6 | 5 | 1 | 4 |

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Report ErrorA compound has an M_{r} of 264 and contains 54.2% oxygen by mass. The number of oxygen atoms in each molecule of this compound is

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Correct, the answer is C!

The oxygen must have a total M_{r} of 264 * 0.542 = 143.088

The number of oxygen atoms that would produce this is 143.088 ÷ 16 = 8.94

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Report ErrorA compound contains 0.3 mol of hydrogen, 0.0600 mol of oxygen and 0.090 mol of nitrogen. The number of H atoms in the empirical formula of this compound is

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Correct, the answer is D!

Calculate the empirical formula:

Element | N | H | O |
---|---|---|---|

Mass/Percentage | - | - | - |

/\ ÷ M_{r} (Moles) |
0.09 | 0.3 | 0.06 |

/\ ÷ Smallest (Ratio) | 1.5 | 5 | 1 |

Final ratio | 3 | 10 | 2 |

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Report ErrorThe mass of hydrogen in grams in C_{6}H_{12}O_{6} that contains 48.0 g of C is

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_{r}. The moles of C atoms is therefore 48 ÷ 12 = 4

The moles of C_{6}H_{12}O_{6} is therefore 4 ÷ 6 = 0.666

The moles of H are therefore 0.666 * 12 = 8

The mass of H is therefore 8 * 1 = 8

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Report ErrorWhich one of the answers has the highest percentage N by mass?

Select an answer from the options

Correct, the answer is C!

The M_{r} for A is 114, of which N accounts for 14. So (14 ÷ 114) * 100 = 12.28%

Repeat for the others

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Report ErrorA compound of M_{r} 160 contains 0.175 mol of C, 0.140 mol of H and 0.0350 mol of N. How many C atoms would be found in the empirical and molecular formula?

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Correct, the answer is C!

Calculate the empirical formula, which is C_{5}H_{4}N, as below:

Element | C | H | N |
---|---|---|---|

Mass/Percentage | - | - | - |

/\ ÷ M_{r} (Moles) |
0.175 | 0.140 | 0.0350 |

/\ ÷ Smallest (Ratio) | 5 | 4 | 1 |

Final ratio | 5 | 4 | 1 |

The empirical mass of this formula is 78. The molecular formula is therefore double the empirical formula (C_{10}H_{8}N_{2})

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Report ErrorIf a compound consists of Na_{3} (PO_{4})_{2} and Na(OH)_{2} and a molar ratio of sodium to phosphorus of 5:3, which formula would fit this information?

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Correct, the answer is B!

This simply means there must be a ratio of 5 Na atoms to every 3 P atoms. The total number of atoms for each must therefore be in this ratio

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Report ErrorAmmonia is manufactured as shown below:

N_{2}_{(g)} + 3H_{2}_{(g)} → 2NH_{3}_{(g)}

If 28.0 g of N_{2} produces 17.0 g of NH_{3}. What is the percentage yield?

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Correct, the answer is A!

Moles = Mass ÷ M_{r}. So, moles of N_{2} is 28 ÷ 28 = 1

N_{2} and NH_{3} are in a 1:2 ratio, so moles of NH_{3} is 1 * 2 = 2

Mass = Moles * M_{r}. So, mass of NH_{3} is therefore 2 * 17 = 34

17 as a percentage of 34 is (17 ÷ 34) * 100 = 50%

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Report Error10 g of calcium is combined with bromine to form CaBr_{2} . What mass of CaBr_{2} (M_{r} = 200) would be formed if the actual yield is 50 % compared to the theoretical yield?

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Correct, the answer is B!

The equation would be: Ca + Br_{2} → CaBr_{2}

Moles = Mass ÷ M_{r}. So, moles of calcium is 10 ÷ 40 = 0.25

Calcium and CaBr_{2} are in a 1:1 ratio, so moles of CaBr_{2} is 0.25

Mass = Moles * M_{r}. So, mass of CaBr_{2} is therefore 0.25 * 200 = 50

50 is the theoretical yield, so the actual yield would be 50% of this. So, 50 * 0.5 = 25

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Report ErrorIf 10 g of Al produced 1 g of H_{2} as per the equation below, the yield of H_{2} as a % would be

2Al + 3H_{2}SO_{4} → 3H_{2} + Al_{2}(SO_{4})_{3}

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Correct, the answer is A!

Moles = Mass ÷ M_{r}. So, moles of Al is 10 ÷ 27 = 0.37

Al and H_{2} are in a 2:3 ratio, so moles of H_{2} is 0.37 ÷ 2 * 3 = 0.55

Mass = Moles * M_{r}. So, mass of H_{2} is therefore 0.55 * 2 = 1.11

1 as a percentage of 1.11 is (1 ÷ 1.11) * 100 = 90%

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Report ErrorWhen 200 g of FeS_{2} are reacted with excess oxygen, 112.0 g of Fe_{2}O_{3} are formed alongside sulfur dioxide. By writing a balanced equation, the percentage yield of Fe_{2}O_{3} could be calculated as

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Correct, the answer is C!

The equation would be: 4FeS_{2} + 7O_{2} → 2Fe_{2}O_{3} + 4SO_{2}

Moles = Mass ÷ M_{r}. So, moles of FeS_{2} is 200 ÷ 120 = 1.66

FeS_{2} and Fe_{2}O_{3} are in a 4:2 ratio, so moles of Fe_{2}O_{3} is 1.66 ÷ 4 * 2 = 0.833

Mass = Moles * M_{r}. So, mass of Fe_{2}O_{3} is therefore 0.833 * 160 = 133.33

112 as a percentage of 133.33 is (112 ÷ 133.33) * 100 = 84%

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Report ErrorWhen 65 g of Fe_{2}O_{3} was added as per the following equation, 38.5 g of Fe was obtained. The percentage yield for this reaction would be

Fe_{2}O_{3} + 3CO → 2Fe + 3CO_{2}

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Correct, the answer is D!

Moles = Mass ÷ M_{r}. So, moles of Fe_{2}O_{3} is 65 ÷ 160 = 0.406

Fe_{2}O_{3} and Fe are in a 1:2 ratio, so moles of Fe is 0.406 * 2 = 0.8125

Mass = Moles * M_{r}. So, mass of Fe is therefore 0.8125 * 56 = 45.5

38.5 as a percentage of 45.5 is (38.5 ÷ 45.5) * 100 = 84.6%

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Report ErrorIf trying to form 28.4 g of P_{4}O_{10}, one would need how many moles of O_{2} ?

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Correct, the answer is C!

The equation would be: 4P + 5O_{2} → P_{4}O_{10}

Moles = Mass ÷ M_{r}. So, moles of P_{4}O_{10} is 28.4 ÷ 284 = 0.1

P_{4}O_{10} and O_{2} are in a 1:5 ratio, so moles of O_{2} is 0.1 * 5 = 0.5

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Report ErrorWhen 10 g of impure MgCO_{3} is heated as below, 0.075 moles of CO_{2} are produced. What is the percentage purity of the MgCO_{3} ? (Assume no impurities produce CO_{2})

MgCO_{3} _{(s)} → MgO_{(s)} + CO_{2} _{(g)}

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Correct, the answer is D!

CO_{2} and MgCO_{3} are in a 1:1 ratio, so moles of pure MgCO_{3} is 0.075

Mass = Moles * M_{r}. So, mass of pure MgCO_{3} is therefore 0.075 * 84 = 6.3

6.3 as a percentage of 10 is (6.3 ÷ 10) * 100 = 63%

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Report ErrorWhen 224 g of carbon monoxide react through the given equation, which of the following is true?

CO_{(g)} + ½O_{2} _{(g)} → CO_{2} _{(g)}

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Correct, the answer is A!

Moles = Mass ÷ M_{r}. So, moles of CO is 224 ÷ 28 = 8. This is also the moles of CO_{2}

CO and O_{2} are in a 1:0.5 ratio, so moles of O_{2} is 8 * 0.5 = 4

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Report ErrorIn the following reaction, with 350 g of each reactant, which one would be the limiting reagent?

Ca_{3}(PO_{4})_{2} + 3SiO_{2} + 5C + 5O_{2} + 3H_{2}O → 3CaSiO_{3} + 5CO_{2} + 2H_{3}PO_{4}

Select an answer from the options

Correct, the answer is B!

Limiting reagents are found by finding moles of each reactant and dividing by the coefficient to get a 1:1 ratio for all. The smallest is limiting. This can be done using a table, like below:

Species | Ca_{3}(PO_{4})_{2} |
SiO_{2} |
C | O_{2} |
H_{2}O |
---|---|---|---|---|---|

Mass | 350 | 350 | 350 | 350 | 350 |

/\ ÷ M_{r} (Moles) |
1.129 | 5.833 | 29.17 | 10.94 | 19.44 |

/\ ÷ Coefficient | 1.129 | 1.944 | 5.834 | 5.47 | 6.48 |

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Report ErrorThe following reaction shows the formation of aspirin from salicylic acid and ethanoic anhydride:

2 C_{7}H_{6}O_{3} + C_{4}H_{6}O_{3} → 2C_{9}H_{8}O_{4} + H_{2}O

The maximum mass of aspirin produced from 2.5 g of salicylic acid (C_{7}H_{6}O_{3}) and excess ethanoic anhydride would be

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_{r}. So, moles of C_{7}H_{6}O_{3} is 2.5 ÷ 138 = 0.0181

C_{7}H_{6}O_{3} and aspirin are in a 2:2 ratio, so moles of aspirin is 0.0181

Mass = Moles * M_{r}. So, mass of aspirin is therefore 0.0181 * 180 = 3.258

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Report ErrorIf 5.0 g of Ca are heated with 5.0 g of S, to completion, in the equation below, then the maximum mass _{(g)} of calcium sulphide that would form is

Ca_{(s)} + S_{(s)} → CaS_{(s)}

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Correct, the answer is D!

Moles = Mass ÷ M_{r}. So, moles of Ca is 5 ÷ 40 = 0.125 and moles of S is 5 ÷ 32 = 0.156

Ca is therefore limiting and so the answer must include 40 but not 32

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Report ErrorIf 10.0 g of phosphorus and 10.0 g of bromine produce 9.6 g of product, which reagent was limiting, and what is the percentage yield for this reaction as shown below?

2P_{(s)} + 3Br_{2} _{(g)} → 2PBr_{3} _{(l)}

Select an answer from the options

Correct, the answer is D!

Moles = Mass ÷ M_{r}. So, moles of P is 10 ÷ 31 = 0.322 and moles of Br is 10 ÷ 160 = 0.0625

Br is therefore limiting

Br and PBr_{3} are found in a 3:2 ratio, so moles of PBr_{3} is 0.0625 ÷ 3 * 2 = 0.04166

Mass = Moles * M_{r}. So, mass of PBr_{3} is therefore 0. 04166 * 271 = 11.2916

9.6 as a percentage of 11.29 is (9.6 ÷ 11.29) * 100 = 85.02%

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Report ErrorIf 6.0 mol of C_{2}H_{3}Cl and 6.0 mol of O_{2} are reacted as below, how many moles of CO_{2} would be made?

C_{2}H_{3}Cl + 2.5O_{2} → 2CO_{2} + H_{2}O + HCl

Select an answer from the options

Correct, the answer is A!

Given C_{2}H_{3}Cl and O_{2} are found in a 1:2.5 ratio, O_{2} would be limiting if there was the same moles

O_{2} and CO_{2} are found in a 2.5:2 ratio, so moles of CO_{2} is 6 ÷ 2.5 * 2 = 4.8

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Report ErrorEqual masses of oxygen and hydrogen gas are reacted in a sealed reaction vessel to produce

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Correct, the answer is A!

The equation would be: O_{2} + 2H_{2} → 2H_{2}O. Let’s assume 1g of each is present

Moles = Mass ÷ M_{r}. So, moles of O_{2} is 1 ÷ 32 = 0.0312 and moles of H_{2} is 1 ÷ 2 = 0.5. So, after dividing by the coefficient, H_{2} is 0.25. O_{2} is therefore limiting

There would therefore be some H_{2} left over in addition to the water produced, but no O_{2}

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Report ErrorCompletely burning 0.60 mol of a hydrocarbon produced CO_{2} and H_{2}O of 1.80 and 2.40 mol respectively. The hydrocarbon must have been

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Correct, the answer is C!

When burning a hydrocarbon, the moles of CO_{2} and H_{2}O produced give the moles of C and H respectively in the compound. So, moles of C is 1.8 and moles of H is 2.4

The ratio of C:H is therefore 1.8:2.4, which simplifies to 3:4

So, the compound would have been C_{3}H_{4}

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Report ErrorThe compound NH_{4}V_{3}O_{8} is found in the following reactions:

2NH_{3}_{(g)} + V_{2}O_{5}_{(s)} + H_{2}O_{(l)} → 2NH_{4}VO_{3} _{(aq)}

3NH_{4}VO_{3} _{(aq)} + 2HCl_{(aq)} → NH_{4}V_{3}O_{8}_{(aq)} + 2NH_{4}Cl_{(aq)} + H_{2}O_{(l)}

If all reactants except ammonia are provided in excess, what is the maximum yield of NH_{4}V_{3}O_{8} producible from ¼ moles of ammonia?

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Correct, the answer is D!

Ammonia and NH_{4}VO_{3} are in a 2:2 ratio, so the moles of NH_{4}VO_{3} would be ¼ in reaction 1

NH_{4}VO_{3} and NH_{4}V_{3}O_{8} are in a 3:1 ratio, so assuming all ¼ moles are used in reaction 2, the moles of NH_{4}V_{3}O_{8} would be ¼ ÷ 3 = 1/12

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Report ErrorA compound of mass 0.496 g consisted of only C, H and O. It was completely burned in oxygen, producing 1.56 g of CO_{2} and 0.638 g of H_{2}O

(a) Calculate the number of moles of carbon dioxide formed [1]

(b) Calculate the number of moles of water formed [1]

(c) What is the empirical formula of the substance? [1]

Write out your answer in the box

(a) Moles = Mass ÷ M_{r}. So, moles of CO_{2} is 1.56 ÷ 44 = 0.0355 mol. ✓

(b) Moles = Mass ÷ M_{r}. So, moles of H_{2}O is 0.638 ÷ 18 = 0.0354 mol. ✓

(c) The ratio of moles of CO_{2} to H_{2}O is 1:1, so the empirical formula would be CH. ✓

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Report Error6.0 dm^{3} of sulfur dioxide and 4.0 dm^{3} of oxygen react as shown below. The volume of sulfur trioxide, in dm^{3}, formed would be

2SO_{2} _{(g)} + O_{2} _{(g)} → 2SO_{3} _{(g)}

Select an answer from the options

Correct, the answer is C!

The ’true’ volume of sulfur dioxide would be 6 ÷ 2 = 3. The volume of O_{2} would still be 4

Therefore, sulfur dioxide is limiting. (Don’t forget to divide by the coefficients for limiting reagents!)

Sulfur dioxide and sulfur trioxide are in 2:2 ratio. So, volume sulfur trioxide would be 6

**Note:** When all regents are in the gaseous state, you can use volumes just like you would moles, saving lots of work!

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Report ErrorChoose the answer with the correct product from the reaction between 400 cm^{3} of H_{2} and 300 cm^{3} of Cl_{2}? Assume constant temperature and pressure

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Correct, the answer is D!

The equation would be: H_{2}_{(g)} + Cl_{2}_{(g)} → 2HCl_{(g)}

Cl_{2} is clearly limiting

Cl_{2} and HCl are in a 1:2 ratio, so volume of HCl would be 300 * 2 = 600

**Note:** H_{2} and Cl_{2} are in a 1:1 ratio, hence the 100cm^{3} of H_{2} left

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Report ErrorAccording to the equation below, what volume of air is required to react with 5 dm^{3} of SO_{2} ? Assume air contains 20% oxygen

2SO_{2} _{(g)} + O_{2} _{(g)} → 2SO_{3} _{(g)}

Select an answer from the options

Correct, the answer is B!

SO_{2} and O_{2} are in a 2:1 ratio, so volume of O_{2} required is 5 ÷ 2 = 2.5

If air contains 20% O_{2} , the volume of air required would be 2.5 ÷ 0.2 = 12.5

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Report ErrorEqual volumes of H_{2} and O_{2} reacted together will produce

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Correct, the answer is A!

The equation would be: O_{2} + 2H_{2} → 2H_{2}O

Let’s assume 1dm^{3} of each is present

‘True’ volume of O_{2} is still 1 and ‘true’ volume of H_{2} is 1 ÷ 2 = 0.5. (Don’t forget to divide by coefficients for limiting reagents!). So H_{2} is limiting

There would therefore be some O_{2} left over in addition to the water produced, but no H_{2}

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Report ErrorFor the following combustion of a hydrocarbon, if 300 cm^{3} of propane is burnt in excess oxygen, the volume of carbon dioxide produced is:

C_{3}H_{8}_{(g)} + 5O_{2} _{(g)} → 3CO_{2} _{(g)} + 4H_{2}O_{(g)}

Select an answer from the options

Correct, the answer is C!

Propane and CO_{2} are in 1:3 ratio. The volume of CO_{2} is therefore 300 * 3 = 900cm^{3}

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Report ErrorA 600 cm^{3} sample of Xe gas has a pressure of 2.335x10^{4} Pa at 150 K. What pressure will it exert if the volume and temperature are increased to 0.8 dm^{3} and 200 K?

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Correct, the answer is C!

The equation to use here is P1 * V1T1 = P_{2} * V_{2}T2

So, 2.335x10^{4} * 0.6150 = P_{2} * 0.8200. Then rearrange

**Note:** This equation generally requires T in Kelvin, P in Pa and V in dm^{3}. If you want, you can use any units for V and P, but they must be the same on both sides

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Report ErrorThe temperature of 210 cm^{3} H_{2} gas is lowered from 40.0 °C to 0 °C at constant pressure. What is the final volume?

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Correct, the answer is A!

The equation to use here is V1T1 = V_{2}T2

So, 210313 = V_{2}273. Then rearrange

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Report Error4.00 mol of He, at 27 °C and 3.00 atm, would fit in a vessel of volume

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Correct, the answer is D!

The equation to use here is PV=nRT

So, 3x10^{5} * V = 4 * 8.31 * 300. Then rearrange to get V in m^{3} as 0.03324

**Note:** This equation requires P in Pa, V in m^{3} and T in Kelvin without exception!

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Report Error0.673 g of a substance has a volume of 272 cm^{3} at 42 °C and 103.4 kPa. What is its molar mass?

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Correct, the answer is A!

The equation to use here is PV=nRT

So, 103.4x10^{3} * 2.72x10^{-4} = n * 8.31 * 315. Then rearrange to get moles as 0.0107.
M_{r} = Mass ÷ Moles

So, 0.673 ÷ 0.0107 = 62.64

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Report ErrorA gas of density 2.48 g dm^{-3} at 40 °C and 1.3 atm would have a molar mass closest to which answer?

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Correct, the answer is B!

The equations to use here are M_{r} = MassPV ÷ RT and Density = Mass ÷ Volume

So, M_{r} = Mass(1.3x10^{5} * V) ÷ (8.31 * 313)

This simplifies to M_{r} = Mass49.98 * V. So, 49.98 * M_{r} = MassV

Therefore, 49.98 * M_{r} = 2.48x10^{3}. (Don’t forget to convert g/dm^{3} into g/m^{3}!)

Then rearrange to get 49.62

**Note:** This question is tough!

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Report ErrorBromine can be prepared by the reaction of hydrogen bromide with manganese(IV) oxide as shown:

4HBr_{(aq)} + MnO_{2} _{(s)} → Br_{2} _{(g)} + MnBr_{2} _{(aq)} + 2H_{2}O_{(l)}

If 17 g of HBr reacted completely with manganese(IV) oxide

(a) Calculate how many moles of hydrogen bromide reacted. [1]

(b) Calculate the volume of bromine gas produced (at STP). [2]

Write out your answer in the box

(a) Moles = Mass ÷ M_{r}. So, moles is 17 ÷ 79.9 = 0.2128. ✓

(b) HBr and Br_{2} are in 4:1 ratio, so moles of Br_{2} is 0.2128 ÷ 4 = 0.0532. ✓

Volume = Moles * 22.7. So, volume is 0.0532 * 22.7 = 1.207dm^{3} → 1.21dm^{3} (3sf). ✓

**Note:** STP uses 22.7 and RTP uses 24dm^{3}!

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Report ErrorIf 42 g of C_{3}H_{6} undergoes combustion as shown, calculate the volume of carbon dioxide produced at STP:

2C_{3}H_{6}_{(g)} + 9O_{2} _{(g)} → 6CO_{2} _{(g)} + 6H_{2}O_{(l)}

Write out your answer in the box

Moles = Mass ÷ M_{r}. So, moles of C_{3}H_{6} is 42 ÷ 42 = 1

C_{3}H_{6} and CO_{2} are in 2:6 ratio, so moles of CO_{2} is 1 ÷ 2 * 6 = 3. ✓

Volume = Moles * 22.7. So, volume is 3 * 22.7 = 68.1dm^{3}. ✓

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Report ErrorExcess HCl reacts with Be to produce 50 cm^{3} of hydrogen gas at 20 °C and 1.08x10^{4} Pa. The equation for this reaction is shown below:

Be + 2HCl → BeCl_{2} + H_{2}

(a) Calculate the moles of hydrogen gas given off. [2]

(b) Calculate the mass of Be that was required for this reaction. [2]

Write out your answer in the box

(a) n = PVRT. ✓

So, moles of H_{2} is (1.08x10^{4} * 5x10^{-5})(8.31 * 293) = 2.22x10^{-4}. ✓

(b) H_{2} and Be are in 1:1 ratio, so moles of Be is 2.22x10^{-4}. ✓

Mass = Moles * M_{r}. So, mass is 2.22x10^{-4} * 9 = 0.001996 → 0.00200 (3sf). ✓

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Report ErrorWhen manganese(IV) oxide (MnO_{2}) is heated it can decompose. Heating 5.00 g of MnO_{2} would produce what volume of oxygen in dm^{3}? (18 °C and 1.05x10^{5} Pa)

3MnO_{2} → Mn3O_{4} + O_{2}

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Moles = Mass ÷ M_{r}. So, moles of MnO_{2} is 5 ÷ 87 = 0.0574. ✓

MnO_{2} and O_{2} are found in 3:1 ratio. So, moles of O_{2} is 0.0574 ÷ 3 = 0.01915. ✓

V = nRTP. So, volume of O_{2} is (0.01915 * 8.31 * 291)1.05x10^{5} = 4.41x10^{-4} m^{3} ✓

Therefore, V in dm^{3} is 4.41x10^{-4} * 1000 = 0.441dm^{3} (3sf). ✓

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Report Error1.01 g of KNO_{3} is dissolved in water to make 0.500 dm^{3} of solution. What would be the concentration in mol dm^{-3}?

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Correct, the answer is B!

Moles = Mass ÷ M_{r}. So, moles of KNO_{3} is 1.01 ÷ 101 = 0.01

Concentration = Moles ÷ Volume. So, concentration is 0.01 ÷ 0.5 = 0.02

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Report ErrorWhat volume in dm^{3} of 0.6 mol dm^{-3} NaCl solution can be prepared from 0.12 mol of NaCl?

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Correct, the answer is B!

Volume = Moles ÷ Concentration. So, volume is 0.12 ÷ 0.6 = 0.2

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Report ErrorHow many moles of HBr are in 25 cm^{3} of 0.2 mol dm^{-3} hydrogen bromide?

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Correct, the answer is A!

Moles = Volume * Concentration. So, moles is 0.025 * 0.2 = 0.005

**Note:** Volume must be in dm^{3} for any mole calculation!

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Report ErrorThe mass of ammonium nitrate required make a solution of volume 500 cm^{3} and concentration 0.40 mol dm^{-3} is

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Correct, the answer is D!

Moles = Volume * Concentration. So, moles is 0.5 * 0.4 = 0.2

Mass = Moles * M_{r}. So, mass of ammonium nitrate is 0.2 * 80 = 16

**Note:** Ammonium nitrate is NH_{4}NO_{3} – useful to learn

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Report ErrorIf a 150 cm^{3} sample of vinegar has a concentration of 0.25 mol dm^{-3} of ethanoic acid, how many moles of ethanoic acid are present?

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Correct, the answer is B!

Moles = Volume * Concentration. But 150cm^{3} needs to be converted to dm^{3} by ÷ 1000

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Report ErrorA sample of HNO_{3} _{(aq)} of volume 20.0 cm^{3} and concentration 0.400 mol dm^{-3} is made into NaNO_{3} _{(aq)}. What volume in cm^{3} of 0.200 mol dm^{-3} NaOH_{(aq)} would be required?

Select an answer from the options

Correct, the answer is C!

The equation would be: HNO_{3} _{(aq)} + NaOH_{(aq)} → NaNO_{3} _{(aq)} + H_{2}O_{(l)}

The equation to use here is C_{1} * V_{1}n_{1} = C_{2} * V_{2}n_{2}

So, 0.4 * 201 = 0.2 * V_{2}1. Then rearrange

**Note:** With the C_{1} * V_{1}n_{1} = C_{2} * V_{2}n_{2} equation the units can be anything, as long as they are
the same on both sides. n stands for the coefficient in the equation relating to that species

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Report ErrorHow much 0.100 mol dm^{-3} NaOH in cm^{3} is required for full titration with 10.0 cm^{3} of 0.050 mol dm^{-3} H_{2}SO_{4} ?

Select an answer from the options

Correct, the answer is C!

The equation would be: H_{2}SO_{4} + 2NaOH → Na_{2}SO_{4} + 2H_{2}O

The equation to use here is C_{1} * V_{1}n_{1} = C_{2} * V_{2}n_{2}

So, 0.1 * V12 = 0.05 * 101. Then rearrange

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Report Error50.0 cm^{3} of sulphuric acid neutralises 36.2 cm^{3} of 0.225 mol dm^{-3} NaOH. The concentration of the H_{2}SO_{4} is thus

Select an answer from the options

Correct, the answer is C!

The equation would be: H_{2}SO_{4} + 2NaOH → Na_{2}SO_{4} + 2H_{2}O

The equation to use here is (C_{1} * V_{1}) ÷ n_{1} = (C_{2} * V_{2}) ÷ n_{2}

So, (0.225 * 36.2) ÷ 2 = (C_{2} * 50) ÷ 1

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Report ErrorIn the below reaction, how much AgF would be precipitated if 45.0 cm^{3} of 0.300 M KF solution was added to excess silver nitrate?

AgNO_{3} _{(aq)} + KF_{(aq)} → AgF_{(s)} + KNO_{3} _{(aq)}

Select an answer from the options

Correct, the answer is B!

Moles = Concentration * Volume. So, moles of KF is 0.3 * 0.045 = 0.0135

KF and AgF are in 1:1 ratio, so moles of AgF = 0.0135

Mass = Moles * M_{r}. So, mass of AgF is 0.0135 * 127 = 1.71

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Report ErrorThe concentration of sulphate ions in 500.0 cm^{3} of 0.30 mol dm^{-3} Fe_{2}(SO_{4})_{3} solution would be

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Correct, the answer is D!

There are three sulphate ions per molecule of Fe_{2}(SO_{4})_{3}, so the concentration is 0.90 mol dm^{-3}

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Report ErrorWhat volume of 3.0 M nitric acid is needed to react 2.83 g of zinc according to the equation shown?

3Zn_{(s)} + 8HNO_{3} _{(aq)} → 3Zn(NO_{3})_{2}_{(aq)} + 2NO_{(g)} + 4H_{2}O_{(l)}

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_{r}. So, moles of zinc is 2.83 ÷ 65 = 0.0435

Zinc and nitric acids are in a 3:8 ratio. So, moles of nitric acid is 0.0435 ÷ 3 * 8 = 0.116

Volume = Moles ÷ Concentration. So, volume is 0.116 ÷ 3 = 0.0386

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Report ErrorWhat is the concentration of Li^{+} ions resulting from 5 g of Li_{2}CO_{3} •7H_{2}O being dissolved in water of 250 cm^{3}?

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_{r}. So, moles of Li_{2}CO_{3} •7H_{2}O is 5 ÷ 200

There are 2 Li atoms in each molecule, so (5 ÷ 200) * 2 gives the moles of Li^{+} atom

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Report ErrorA student determined the concentration of HCl_{(aq)} by titration. When 90.0 cm^{3} of acid was titrated to end point, 18.0 cm^{3} of 0.18 M LiOH was required. What was the molarity of the HCl solution?

Select an answer from the options

Correct, the answer is C!

The equation would be: HCl + LiOH → LiCl + H_{2}O

The equation to use here is C_{1} * V_{1}n_{1} = C_{2} * V_{2}n_{2}

So, 18 * 0.0181 = C_{2} * 901. Then rearrange

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Report ErrorManganese(II) hydroxide is produced by reacting manganese(II) sulphate with potassium hydroxide as shown below. How much could be produced if 15 cm^{3} of 0.2 mol dm^{-3} potassium hydroxide is added to 40 cm^{3} 0.15 mol dm^{-3} manganese sulphate?

MnSO_{4} _{(aq)} + 2KOH_{(aq)} → Mn(OH)_{2}_{(s)} + K_{2}SO_{4} _{(aq)}

Select an answer from the options

Correct, the answer is A!

As we have information on both reactants, we need to establish which is limiting

Moles = Mass ÷ M_{r}. So, moles of KOH is 0.015 * 0.2. = 0.003. Moles of MnSO_{4} is 0.040 * 0.15 = 0.006

The ‘true’ moles of KOH is 0.003 ÷ 2 = 0.0015, so KOH is limiting. (Coefficient of 2)

As KOH and Mn(OH)_{2} are in 2:1 ratio, moles of Mn(OH)_{2} is therefore 0.003 ÷ 2 = 0.0015

Mass = Moles * M_{r}. So, mass is 0.0015 * 89 = 0.1335

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Report ErrorWhat mass of silver bromide can be produced from 25 cm^{3} of 0.17 mol dm^{-3} sodium bromide and an excess of aqueous silver ions?

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Correct, the answer is B!

The equation would be: Ag^{+} + NaBr → AgBr + Na^{+}

Moles = Volume * Concentration. So, moles of sodium bromide is 0.025 * 0.17

Sodium bromide and silver bromide are in 1:1 ratio, so moles of silver bromide is 0.025 * 0.17

Mass = Moles * M_{r}. So, mass is (0.025 * 0.17) * (107.87 + 79.9)

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Report ErrorBy writing an equation, calculate the concentration of NaOH if 70.00 cm^{3} successfully neutralises 50.00 cm^{3} of 0.3000 mol dm^{-3} H_{2}SO_{4}

Select an answer from the options

Correct, the answer is B!

The equation would be: H_{2}SO_{4} + 2NaOH → Na_{2}SO_{4} + 2H_{2}O

The equation to use here is C_{1} * V_{1}n_{1} = C_{2} * V_{2}n_{2}

So, 0.3 * 501 = C_{2} * 702. Then rearrange

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Report ErrorPotassium hydroxide reacts with phosphoric(V) acid as shown below:

3KOH + H_{3}PO_{4} → K_{3}PO_{4} + 3H_{2}O

If 50.00 cm^{3} of 0.20 mol dm^{-3} KOH reacts with 0.04 mol dm^{-3} H_{3}PO_{4} , the volume of H_{3}PO_{4} , in cm^{3}, necessary would be calculated by which of the following?

Select an answer from the options

Correct, the answer is D!

The equation to use here is C_{1} * V_{1}n_{1} = C_{2} * V_{2}n_{2}

So, 0.2 * 503 = 0.04 * V_{2}1

Multiplying by 0.2 is the same as dividing by 5. Then rearrange

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Report ErrorWhat volume of 0.500 M HF is required to react with 9.3 g of calcium carbonate according to the equation:

CaCO_{3} _{(s)} + 2HF_{(aq)} → CaF_{2}_{(aq)} + H_{2}O_{(l)} + CO_{2} _{(g)}

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_{r}. So, moles of calcium carbonate is 9.3 ÷ 100 = 0.093

Calcium carbonate and HF are in 1:2 ratio. So, moles of HF is 0.093 * 2 = 0.186

Volume = Moles ÷ Concentration. So, volume is 0.186 ÷ 0.5 = 0.372dm^{3}

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Report Error60 cm^{3} of HCl of concentration 1 M is added to 40 cm^{3} of 1 M NaOH. The number of moles of HCl left unreacted would be given by which answer?

Select an answer from the options

Correct, the answer is C!

The equation would be: HCl + NaOH → NaCl + H_{2}O

Since they react in a 1:1 ratio, all NaOH would be used up and 20cm^{3} of HCl would be left

Moles = Concentration * Volume. So, moles is 0.02 * 1 = 0.02

**Note:** Don’t forget to convert cm^{3} to dm^{3}!

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Report ErrorWhat mass of PbCl_{2} would be precipitated if excess Pb(NO_{3})_{2} solution were combined with 87.0 cm^{3} of 0.100 M KCl solution?

Select an answer from the options

Correct, the answer is B!

The equation would be: Pb(NO_{3})_{2} + 2KCl → PbCl_{2} + 2KNO_{3}

Moles = Concentration * Volume. So, moles of KCl is 0.1 * 0.087 = 0.0087

KCl and PbCl_{2} are in 2:1 ratio, so moles of PbCl_{2} is 0.0087 ÷ 2 = 0.00435

Mass = Moles * M_{r}. So, mas is 0.00435 * 278 = 1.209

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Report Error50.00 cm^{3} of silver nitrate solution is reacted with 100.00 cm^{3} of calcium bromide solution. The concentrations of these solutions respectively could be:

2AgNO_{3} + CaBr_{2} → 2AgBr + Ca(NO_{3})_{2}

Select an answer from the options

Correct, the answer is D!

Moles = Concentration * Volume. So, using the coefficients as moles, we can say for silver nitrate 2 = C1 * 50 and for calcium bromide 1 = C_{2} * 100

Therefore, C1 = 0.04 and C_{2} = 0.01. This is a ratio of 4:1, so, the answer would match this ratio

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Report Error0.05 mol of the iodide of element A was dissolved in water to make a total volume of 1 dm^{3}. 25 cm^{3} of this solution reacted with 25 cm^{3} of 0.2 mol dm^{-3} silver nitrate solution. The formula of the iodide is

Select an answer from the options

Correct, the answer is D!

Since we don’t know element A, the equation would be: AIx + AgNO_{3} → IxNO_{3} + AgI where x denotes an unknown number

Moles = Concentration * Volume. So, moles of silver nitrate is 0.2 * 0.025 = 0.005.
Assuming a 1:1 ratio, we can therefore say 0.005 moles of AIx was required

Since we used a 25cm^{3} sample, which is 1/40th of the total 1dm^{3}, the total moles of AIx that would be required for this reaction is 0.005 * 40 = 0.2

This 0.2 mol is 4x the amount of moles of iodide A that were prepared, so the equation must have been 4x our assumed 1:1 ratio: AI_{4} + AgNO_{3} → I_{4}NO_{3} + AgI

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Report ErrorA 25 cm^{3} sample of a solution which is 0.05 M in HCl is combined with 10 cm^{3} of 0.05 M NaOH. The concentration of HCl remaining after reaction is given by

Select an answer from the options

Correct, the answer is A!

The equation would be: HCl + NaOH → NaCl + H_{2}O

Moles = Concentration * Volume. So, the moles of HCl left would be (0.05 * 25) - (0.05 * 10)

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Report Error20 cm^{3} of barium fluoride solution, BaF_{2}_{(aq)}, completely reacted with 40 cm^{3} of iron sulphate solution, Fe_{2}(SO_{4})_{3}_{(aq)} to form 1.398 g of BaSO_{4} precipitate. The concentration of the iron sulphate solution is

Select an answer from the options

Correct, the answer is B!

The equation would be: 3BaF_{2} + Fe_{2}(SO_{4})_{3} → 3BaSO_{4} + 2FeF_{3}

Moles = Mass ÷ M_{r}. So, moles of BaSO_{4} is 1.398 ÷ 233 = 0.006

BaSO_{4} and iron sulphate are in 3:1 ratio, so moles of iron sulphate is 0.006 ÷ 3 = 0.002

Concentration = Moles ÷ Volume. So, concentration is 0.002 ÷ 0.04 = 0.05

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Report Error If 0.001 mole of a bromide required 40 cm^{3} of 0.05 M silver nitrate (AgNO_{3}) for reaction. The original bromide could be which of the following?

Select an answer from the options

Correct, the answer is B!

The equation would be: XBr + AgNO_{3} → XNO_{3} + AgBr

Moles = Concentration * Volume. So, moles of AgNO_{3} is 0.05 * 0.04 = 0.002

This is 2x the moles of bromide, so the equation becomes XBr_{2} + 2AgNO_{3} → X(NO_{3})_{2} + 2AgBr

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Report ErrorIn a NaOH and HCl titration, 100 cm^{3} of 0.2 M NaOH is added to 50 cm^{3} of 0.1 M HCl. The molarity of NaOH remaining is thus

Select an answer from the options

Correct, the answer is B!

The equation would be: HCl + NaOH → NaCl + H_{2}O

Moles = Concentration * Volume. So, the moles of NaOH left would be (0.2 * 100) - (0.1 * 50)

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Report ErrorHow much AgCl is precipitated from 0.002 m^{3} of 0.1000 M AgNO_{3} reacting with excess MgCl_{2}?

Select an answer from the options

Correct, the answer is A!

The equation would be: MgCl_{2} + 2AgNO_{3} → 2AgCl + Mg(NO_{3})_{2}

Moles = Concentration * Volume. So, the moles of AgNO_{3} is 0.1 * 2 = 0.2

AgNO_{3} and AgCl are in 2:2 ratio, so moles of AgCl is 0.2

Mass = Moles * M_{r}. So, mass is 0.2 * 143.5 = 28.7

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Report ErrorWhat mass of PbF_{2} can be precipitated from 100 cm^{3} of 0.300 M Pb(NO_{3})_{2} and 100 cm^{3} of 0.200 M NaF?

Select an answer from the options

Correct, the answer is B!

The equation would be: Pb(NO_{3})_{2} + 2NaF → PbF_{2} + 2NaNO_{3}

None of the answers use 0.3, so we assume NaF is limiting

Moles = Concentration * Volume. So, the moles of NaF is 0.2 * 0.1. Check the answers

**Note:** Remember its multiple choice. For speed, work out a partial answer and then scan the options!

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Report ErrorTo discover the percentage of calcium carbonate in 1.216g of clean, dry coral, an experiment was conducted. The coral was added to 10.0 cm^{3} of 5.00 mol dm^{-3} HCl to dissolve. The solution created was made up to 25.0 cm^{3} with distilled water. This diluted solution required 28.0 cm^{3} of 1.00 mol dm^{-3} NaOH for neutralisation. An equation for the initial reaction is shown below:

CaCO_{3} _{(s)} + 2HCl_{(aq)} → CaCl_{2}_{(aq)} + CO_{2} _{(g)} + H_{2}O_{(l)}

(a) Calculate the number of moles of NaOH added. [1]

(b) How many moles of HCl remained in the beaker after the coral was dissolved? [1]

(c) Using your answer to part b), how many moles of acid reacted with the coral? [1]

(d) What mass of CaCO_{3} was present in the coral? [1]

(e) What was the % composition of CaCO_{3} in the coral? [1]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 1 * 0.028 = 0.028 mol. ✓

(b) The equation would be: HCl + NaOH → NaCl + H_{2}O. So, if HCl and NaOH react in 1:1 ratio, the moles of HCl would be the same as NaOH, so = 0.028 mol. ✓

(c) Moles = Concentration * Volume. So, moles of HCl added is 5 * 0.01 = 0.05

If 0.028 remained, moles reacted is 0.05 - 0.028 = 0.022 mol. ✓

(d) HCl and CaCO_{3} are in 2:1 ratio, so moles CaCO_{3} is 0.022 ÷ 2 = 0.011

Mass = Moles * M_{r}. So, mass is 0.011 * 100 = 1.1 g. ✓

(e) 1.1 as a percentage of 1.216 is (1.1 ÷ 1.216) * 100 = 90.4605 → 90.5% (3sf). ✓

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Report Error2.447 g of an ammonia-containing surface cleaner was diluted to 20.0 cm^{3}. This solution required 28.51 cm^{3} of 0.4040 mol dm^{-3} sulphuric acid for neutralisation as per the equation shown:

2NH_{3}_{(aq)} + H_{2}SO_{4} _{(aq)} → (NH_{4})_{2}SO_{4} _{(aq)}

(a) Calculate the number of moles of H_{2}SO_{4} required for this reaction. [1]

(b) Calculate the mass of ammonia present in the surface cleaner. [2]

(c) Show by calculation the percentage by mass of ammonia present in the household cleaner. [1]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 0.404 * 0.02851 = 0.0115 mol. ✓

(b) H_{2}SO_{4} and ammonia are in 1:2 ratio, so moles ammonia is 0.0115 * 2 = 0.023

Mass = Moles * M_{r}. So, mass is 0.023 * 17 = 0.391 g. ✓

(c) 0.3916 as a percentage of 2.447 is (0.391 ÷ 2.447) * 100 = 15.979 → 16.0% (3sf). ✓

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Report Error1.377 g of anhydrous potassium carbonate was made up to 500 cm^{3} using distilled water. 50.00 cm^{3} of this solution was titrated with CH_{3}COOH using methyl orange indicator. The titration required 45.3 cm^{3} of the acid for complete neutralisation.

(a) Write a balanced equation for this neutralisation reaction. [2]

(b) Determine the concentration of the potassium carbonate solution. [2]

(c) Calculate the concentration of the ethanoic acid solution. [2]

Write out your answer in the box

(a) K_{2}CO_{3} + 2CH_{3}COOH → 2CH_{3}COOK + H_{2}CO_{3} . ✓ ✓ (1 mark for balancing)

(b) Moles = Mass ÷ M_{r}. So, moles of potassium carbonate is 1.377 ÷ 138 = 0.009978. ✓

Concentration = Moles ÷ Volume. So, 0.009978 ÷ 0.5 = 0.0200 mol/dm^{3}. ✓

(c) K_{2}CO_{3} and ethanoic acid are in 1:2 ratio, so, moles of ethanoic acid is 0.009978 * 2 = 0.019956. ✓

Concentration = Moles ÷ Volume. So, 0.019956 ÷ 0.0453 = 0.4405 → 0.441 mol/dm^{3} (3sf). ✓

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Report ErrorConsider the following reaction. 0.0487 mol of zinc(II) oxide was added to 39.0 cm^{3} of 2.42 mol dm^{-3} nitric acid solution

ZnO + 2HNO_{3} → Zn(NO_{3})_{2} + H_{2}O

(a) Calculate the amount (in mol) of nitric acid. [1]

(b) Which is the limiting reactant? [1]

(c) Determine the number of moles of zinc(II) nitrate formed. [1]

(d) The product of this reaction was collected as zinc(II) nitrate trihydrate (Zn(NO_{3})_{2}•3H_{2}O). Calculate the mass of this product formed. [2]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 2.42 * 0.039 = 0.0944 mol. ✓

(b) We need to compare in 1:1 ratio, so ‘true’ moles of nitic acid is 0.09438 ÷ 2 = 0.04719

Moles of zinc oxide is 0.0487, so nitric acid is limiting. ✓

(c) Nitric acid and zinc nitrate are in 2:1 ratio, so, moles of zinc nitrate is 0.0944 ÷ 2 = 0.0472 mol. ✓

(d) Mass = Moles * M_{r}. So, mass is 0.0472 * 243 = 11.4696 → 11.5g (3sf). ✓ ✓ (1 mark for 243)

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Report Error16.20x10^{-2} dm^{3} of 0.304 mol dm^{-3} aqueous AgNO_{3} is added to 11.60x10^{-2} dm^{3} of 0.4550 mol dm^{-3} aqueous NaF as per the equation below:

AgNO_{3} _{(aq)} + NaF_{(aq)} → AgF_{(s)} + NaNO_{3} _{(aq)}

(a) Calculate the mass in grams of AgF obtained in this reaction. [4]

(b) By what percentage, more or less, of AgF would the other product produce? [3]

Write out your answer in the box

(a) We need to find out limiting reagent, so, Moles = Concentration * Volume

Moles of AgNO_{3} is 0.304 * 16.20x10^{-2} = 0.049248 ✓ and NaF is 0.4550 * 11.60x10^{-2} = 0.05278. ✓

Therefore, AgNO_{3} is limiting

AgNO_{3} and AgF are in 1:1 ratio, so, moles of AgF = 0.049248. ✓

Mass = Moles * M_{r}. So, mass is 0.049248 * 127 = 6.2545 → 6.25g (3sf). ✓

(b) Using NaF, the moles of AgF = 0.05278. ✓

Mass = Moles * M_{r}. So, mass is 0.05278 * 127 = 6.7036. ✓

The percentage increase is therefore 6.7036 - 6.25446.2544 * 100 = 7.1735 → 7.17% increase (3sf). ✓

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Report ErrorAqueous XO_{4} ^{3-} ions form a precipitate with aqueous rubidium ions, Rb^{+}:

3Rb^{+}_{(aq)} + XO_{4} ^{3-}_{(aq)} → Rb_{3} XO_{4} _{(s)}

When 82.36 cm^{3} of 0.4080 mol dm^{-3} aqueous rubidium ions is added to XO_{4} ^{3-} ions, 4.336 g of precipitate is formed

(a) Calculate the amount (in moles) of Rb^{+} ions used. [1]

(b) Calculate the amount (in moles) of the precipitate formed. [1]

(c) Calculate the M_{r} of this precipitate. [1]

(d) Determine the A_{r} of X and identify the element this represents [2]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 0.408 * 0.08236 = 0.03360 mol. ✓

(b) Rb^{+} and the precipitate are in 3:1 ratio, so, moles of precipitate is 0.0336 ÷ 3 = 0.01120 mol. ✓

(c) M_{r} = Mass ÷ Moles. So, M_{r} is 4.336 ÷ 0.0112336 = 387.1 g/mol. ✓

(d) M_{r} of Rb_{3} XO_{4} is 387.1. So, X is 387.1 – (3 * 85.5) – ( 4 * 16) = 66.6. ✓

The closest element is Zn. ✓

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Report ErrorA dye contains only the elements C, N, H and O. 2.036 g of the dye was oxidised. In doing so, 5.470 g of CO_{2} and 0.697 g of H_{2}O were produced

(a) Calculate the percentage by mass of C and H in the dye. [4]

(b) The % by mass of N in the dye is 10.75 %. Find the empirical formula. [3]

(c) Knowing the M_{r} is 260 g mol^{-1}, find the molecular formula of the dye. [1]

Write out your answer in the box

(a) Moles = Mass ÷ M_{r}. So, moles of CO_{2} = 5.470 ÷ 44 = 0.124318. This is the moles of C in the dye

For H_{2}O, moles is 0.697 ÷ 18 = 0.03872. The moles of H in the dye is thus 0.03872 * 2 = 0.07744. ✓

Mass = Moles * M_{r}. So, mass of C is 0.124318 * 12 = 1.4918g and H is 0.07744 * 1 = 0.07744g. ✓

1.4918 as a percentage of 2.036 is (1.4918÷ 2.036) * 100 = 73.271 → C = 73.3% (3sf) ✓

0.07744 as a percentage of 2.036 is (0.07744 ÷ 2.036) * 100 = 3.803 → H = 3.80% (3sf) ✓

(b) The % of O would be 100 – 73.3 – 3.80 – 10.75 = 12.15%. ✓

Then find the empirical formula as below: ✓

Element | C | N | H | O |
---|---|---|---|---|

Mass/Percentage | 73.3 | 10.75 | 3.8 | 12.15 |

/\ ÷ M_{r} (Moles) |
6.1083 | 0.7678 | 3.8 | 0.7593 |

/\ ÷ Smallest (Ratio) | 8.04 | 1.011 | 5 | 1 |

Final ratio | 8 | 1 | 5 | 1 |

The empirical formula is C_{8}NH_{5}O. ✓

(c) The empirical mass is 131. The M_{r} is twice this, so the formula is C_{16} N_{2}H_{10}O_{2} . ✓

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Report Error98.98 g of hydrated lithium carbonate crystals, Li_{2}CO_{3} •xH_{2}O , were made up to 0.5000 dm^{3}. 12.500 cm^{3} of this solution was neutralized with 24.40 cm^{3} of 0.2000 mol dm^{-3} HCl as shown:

Li_{2}CO_{3} _{(aq)} + 2HCl_{(aq)} → 2LiCl_{(aq)} + CO_{2} _{(g)} + H_{2}O_{(l)}

(a) Calculate the concentration of the lithium carbonate solution. [3]

(b) Find the mass of lithium carbonate present in the 0.500 dm^{3} solution. [2]

(c) Calculate the mass of H_{2}O in the hydrated crystals and hence the value of x. [4]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles of HCl is 0.2 * 0.02440 = 0.00488. ✓

HCl and Li_{2}CO_{3} are in 2:1 ratio, so moles of Li_{2}CO_{3} is 0.00488 ÷ 2 = 0.00244. ✓

Concentration = Moles ÷ Volume. So, concentration is 0.00244 ÷ 0.0125 = 0.1952 mol/dm^{3}. ✓

(b) 0.5 dm^{3} of solution is 4 * 12.5 cm^{3} so the moles of Li_{2}CO_{3} is 0.1952 * 4 = 0.7808. ✓

Mass = Moles * M_{r}. So, mass is 0.7808 * 74 = 57.78g. ✓

(c) Mass of water is 98.98 – 57.78 = 41.2g. ✓

Moles = Mass ÷ M_{r}. So, moles is 41.2 ÷ 18 = 2.2889. ✓

So, moles of water ÷ Li_{2}CO_{3} will give X. So, X is 2.2889 ÷ 0.7808 = 2.93 ✓

X is therefore 3. ✓

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Report Error0.453 g of an alkali metal sulfate (A_{2}SO_{4}) was dissolved in water. Excess barium fluoride solution was then added to precipitate all the sulfate ions as barium sulfate, BaSO_{4} . This precipitate was filtered and dried. It weighed 0.438 g.

(a) Calculate the amount of moles of BaSO_{4} . [1]

(b) Determine the amount (in mol) of A_{2}SO_{4} present. [2]

(c) Determine the molar mass of the alkali metal sulfate. [1]

(d) Deduce the identity of A, showing your workings. [2]

Write out your answer in the box

(a) Moles = Mass ÷ M_{r}. So, moles of BaSO_{4} is 0.438 ÷ 233 = 0.00188 mol. ✓

(b) The equation is A_{2}SO_{4} + BaF_{2} → BaSO_{4} + 2AF ✓

BaSO_{4} and A_{2}SO_{4} are in 1:1 ratio, so moles of A_{2}SO_{4} = 0.00188 mol. ✓

(c) M_{r} = Mass ÷ Moles. So, mass of A_{2}SO_{4} is 0.453 ÷ 0.00188 = 241 g/mol. ✓

(d) M_{r} of A_{2} is 241– 32 – (4 * 16) = 145. ✓

A is therefore 145 ÷ 2 = 72.5 and the closest element is Germanium (Ge). ✓

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Report ErrorAn experiment was conducted to calculate x in Fe(NH_{4})_{2}(SO_{4})_{2}•xH_{2}O. A 3.54 g sample of this compound was dissolved in water and excess BaBr_{2} _{(aq)} was added to precipitate ions. The precipitate containing BaSO_{4} weighed 2.83 g when dried.

(a) Calculate the amount, in moles, of BaSO_{4} in the precipitate. [1]

(b) Calculate the number of moles of sulfate in the 3.54 g of Fe(NH_{4})_{2}(SO_{4})_{2}•xH_{2}O. [2]

(c) State the quantity, in moles, of Fe in the sample. [1]

(d) Determine the mass of the following present in the sample of Fe(NH_{4})_{2}(SO_{4})_{2}•xH_{2}O. [3]

(i) Iron

(ii) Ammonium

(iii) Sulfate

(e) Using your previous answers, determine the number of moles of water present in the 3.54g. [2]

(f) Thus, calculate the value of x. [2]

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(a) Moles = Mass ÷ M_{r}. So, moles of BaSO_{4} is 2.83 ÷ 233 = 0.0121 mol. ✓

(b) The equation is Fe(NH_{4})_{2}(SO_{4})_{2} + 2BaBr_{2} → 2BaSO_{4} + 2Fe(NH_{4})Br_{2} ✓

BaSO_{4} and Fe(NH_{4})_{2}(SO_{4})_{2} are in 2:1 ratio, so moles of Fe(NH_{4})_{2}(SO_{4})_{2} is 0.012145 ÷ 2 = 0.00607.

So, the moles of sulfate is 0.00607 * 2 = 0.0121 mol. ✓

(c) There are twice as many moles of SO_{4} as Fe, so moles of Fe is 0.0121 ÷ 2 = 0.00605 mol. ✓

(d) The moles of ammonium are the same as sulfate. Mass = Moles * M_{r}. So:

Mass of iron is 0.00607 * 56 = 0.339g. ✓

Mass of sulfate is 0.0121 * 96 = 1.16g. ✓

Mass of ammonium is 0.0121 * 19 = 0.230g. ✓

(e) Mass of water is 3.54 - 0.339 - 1.16 - 0.230 = 1.811g. ✓

Moles = Mass ÷ M_{r}. So, moles of water = 1.811 ÷ 18 = 0.101 mol. ✓

(f) Moles of water ÷ Fe(NH_{4})_{2}(SO_{4})_{2} will give X. So, X is 0.101 ÷ 0.00605 = 16.7. ✓

X is therefore 17. ✓ (Allow 16.7)

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Report ErrorTo determine the percentage Cu in 3 samples of brass, each 0.748 g, a three-stage reaction pathway was utilised. This is shown below:

1) Cu_{(s)} + 2HNO_{3} _{(aq)} + 2H^{+}_{(aq)} → Cu^{2+}_{(aq)} + 2NO_{2} _{(g)} + 2H_{2}O_{(l)}

2) 4I^{-}_{(aq)} + 2Cu^{2+}_{(aq)} → 2CuI_{(s)} + I_{2}_{(aq)}

3) I_{2}_{(aq)} + 2S_{2}O_{3} ^{2-}_{(aq)} → 2I^{-}_{(aq)} + S_{4}O_{6}^{2-}_{(aq)}

Titration Repeat | |||
---|---|---|---|

1 | 2 | 3 | |

Initial Volume of S_{2}O_{3} ^{2-} |
73.3 | 10.75 | 3.8 |

Final Volume of S_{2}O_{3} ^{2-} |
6.1083 | 0.7678 | 3.8 |

Volume of S_{2}O_{3} ^{2-} added |
8.04 | 1.011 | 5 |

(a) By using the data in the table and finding an appropriate average, calculate the average number of moles of 0.100 mol/dm^{3} S_{2}O_{3} ^{2-} added in the final step. Assume all volumes are recorded in cm^{3}. [2]

(b) Calculate the average percentage by mass of copper in the three samples of brass. [3]

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(a) Average volume is 38.4 + 38.5 + 38.23 = 38.366. ✓

Moles = Concentration * Volume. So, average moles is 0.1 * 0.038366 = 0.00384 mol. ✓

(b) S_{2}O_{3} ^{2-} and I_{2} are in 2:1 ratio, so, the moles of I_{2} = 0.00384 ÷ 2.
However, I_{2} and Cu^{2+} are in 1:2 ratio, so, moles of Cu^{2+} is (0.00384 ÷ 2) * 2, which = 0.00384.

Since Cu^{2+} and Cu are in 1:1 ratio, moles of Cu = 0.00384. ✓

Mass = Moles * M_{r}. So, mass is 0.00384 * 63.5 = 0.24384g ✓

0.24384 as a percentage of 0.748 is (0.24384 ÷ 0.748) * 100 = 32.5989 → 32.6% (3sf) ✓

**Note:** When working backwards through different reactions you assume all of the reactant comes from the product of the previous equation.

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