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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 A!

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

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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)}

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Correct, the answer is C!

Moles = Mass ÷ Mr. 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 * Mr. 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 * Mr. 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 ÷ Mr. 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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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 * Mr. 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

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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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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 * Mr. So, mass of Al is therefore 0.256 * 27 = 6.917g → 6.92g (3sf) ✓

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Report ErrorCalculate the mass of Fe2O_{3} required to produce 3632 kg of Fe as per the equation shown below:

Fe2O_{3} + 3CO → 2Fe + 3CO_{2}

Write out your answer in the box

Moles = Mass ÷ Mr. So, moles of Fe is 3632 ÷ 56 = 64.857. ✓

Fe and Fe2O_{3} are in a 2:1 ratio, so moles of Fe2O_{3} is 64.857 ÷ 2 = 32.428. ✓

Mass = Moles * Mr. So, mass of Fe2O_{3} is therefore 32.428 * 160 = 5188.57kg → 5190kg (3sf) ✓

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Report ErrorWhich of these options has the greatest mass?

Select an answer from the options

Correct, the answer is D!

Mass = M_{r} * Moles

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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 ÷ Mr. 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 Mr!

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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} * %)[TT_F t=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 ÷ Mr. 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?

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Correct, the answer is B!

Whichever has the fewest moles will have the fewest molecules. Moles = Mass ÷ Mr. 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

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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

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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

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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} = 66.45

So, the mass is 8 * 66.45 = 531.6. 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?

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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 | O | |
---|---|---|---|

Mass/Percentage | 87.5 | 12.5 | 32 |

/\ ÷ M_{r} (Moles) |
6.25 | 12.5 | 2 |

/\ ÷ Smallest (Ratio) | 1 | 2 | 1 |

Final ratio | 1 | 2 | 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 | O | |
---|---|---|---|

Mass/Percentage | 85.7 | 14.3 | 32 |

/\ ÷ M_{r} (Moles) |
7.14 | 14.3 | 2 |

/\ ÷ Smallest (Ratio) | 1 | 2 | 1 |

Final ratio | 1 | 2 | 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

Select an answer from the options

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

Select an answer from the options

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 ÷ Mr. 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 A!

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 formulae 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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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 * Mr. 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}

Select an answer from the options

Correct, the answer is A!

Moles = Mass ÷ Mr. 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 * Mr. 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 Fe2O_{3} are formed alongside sulfur dioxide. By writing a balanced equation, the percentage yield of Fe2O_{3} could be calculated as

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Correct, the answer is C!

The equation would be: 4FeS_{2} + 7O_{2} → 2Fe2O_{3} + 4SO_{2}

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

FeS_{2} and Fe2O_{3} are in a 4:2 ratio, so moles of Fe2O_{3} is 1.66 ÷ 4 * 2 = 0.833

Mass = Moles * Mr. So, mass of Fe2O_{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 Fe2O_{3} was added as per the following equation, 38.5 g of Fe was obtained. The percentage yield for this reaction would be

Fe2O_{3} + 3CO → 2Fe + 3CO_{2}

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Correct, the answer is D!

Moles = Mass ÷ Mr. So, moles of Fe2O_{3} is 65 ÷ 160 = 0.406

Fe2O_{3} and Fe are in a 1:2 ratio, so moles of Fe is 0.406 * 2 = 0.8125

Mass = Moles * Mr. 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}O10, 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}O10

Moles = Mass ÷ Mr. So, moles of P_{4}O10 is 28.4 ÷ 284 = 0.1

P_{4}O10 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 * Mr. 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 ÷ Mr. 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 |
SiO2 | 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} → 2C9H_{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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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}O8 is found in the following reactions:

2NH_{3}_{(g)} + V_{2}O5_{(s)} + H_{2}O_{(l)} → 2NH_{4}VO_{3} _{(aq)}

3NH_{4}VO_{3} _{(aq)} + 2HCl_{(aq)} → NH_{4}V_{3}O8_{(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}O8 producible from ¼ moles of ammonia?

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Correct, the answer is D!

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

NH_{4}V_{3}O and NH_{4}V_{3}O8 are in a 3:1 ratio, so assuming all ¼ moles are used in reaction 2, the moles of NH_{4}V_{3}O8 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 ÷ Mr. So, moles of CO_{2} is 1.56 ÷ 44 = 0.0355 mol. ✓

(b) Moles = Mass ÷ Mr. 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} = ERROR(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 s.t.p). [2]

Write out your answer in the box

(a) Moles = Mass ÷ Mr. 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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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 ÷ Mr. 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

Select an answer from the options

Correct, the answer is D!

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

Mass = Moles * Mr. 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 C1 * V1 n1 = C_{2} * V_{2}n2

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

**Note:** With the C1 * V1n1 = C_{2} * V_{2}n2 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} ?

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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 C1 * V1n1 = C_{2} * V_{2}n2

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

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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 (C1 * V1) ÷ n_{1} = (C_{2} * V_{2}) ÷ n2

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 * Mr. 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} Fe2(SO_{4})_{3} solution would be

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Correct, the answer is D!

There are three sulphate ions per molecule of Fe2(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 ÷ Mr. 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}?

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Correct, the answer is C!

Moles = Mass ÷ Mr. 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 C1 * V1n1 = C_{2} * V_{2}n2

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 ÷ Mr. 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 * Mr. 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 * Mr. 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}

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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 C1 * V1n1 = C_{2} * V_{2}n2

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 C1 * V1n1 = C_{2} * V_{2}n2

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 ÷ Mr. 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?

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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 * Mr. 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

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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, Fe2(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} + Fe2(SO_{4})_{3} → 3BaSO_{4} + 2FeF_{3}

Moles = Mass ÷ Mr. 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?

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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_{2})_{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}?

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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 * Mr. 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?

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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 * Mr. 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})2SO_{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 * Mr. 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]

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

(b) Moles = Mass ÷ Mr. 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]

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(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 * Mr. 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 * Mr. So, mass is 0.049248 * 127 = 6.2545 → 6.25g (3sf). ✓

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

Mass = Moles * Mr. 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 ÷ Mr. 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 * Mr. 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 * Mr. So, mass is 0.7808 * 74 = 57.78g. ✓

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

Moles = Mass ÷ Mr. 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 (A2SO_{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 A2SO_{4} present. [2]

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

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

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

(b) The equation is A2SO_{4} + BaF_{2} → BaSO_{4} + 2AF ✓

BaSO_{4} and A2SO_{4} are in 1:1 ratio, so moles of A2SO_{4} = 0.00188 mol. ✓

(c) M_{r} = Mass ÷ Moles. So, mass of A2SO_{4} is 0.453 ÷ 0.00188 = 241 g/mol. ✓

(d) M_{r} of A2 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 ÷ Mr. 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 * Mr. 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 ÷ Mr. 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]

Write out your answer in the box

(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 * Mr. 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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