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Exam calculations: moles, solutions, titration and yield: GCSE Combined Science Chemistry practice questions

Higher tier content · Grade 8

A second set of questions on exam calculations: moles, solutions, titration and yield. They are different from the ones on the topic page. Try each, then open the answer and compare your working.

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Question 1 Medium

Hydrogen can be made by CH₄ + 2H₂O → CO₂ + 4H₂. Calculate the atom economy for making hydrogen to 3 significant figures. Mr of CH₄ = 16, H₂O = 18 and H₂ = 2.

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Answer: 15.4%

  1. Useful product: 4 × 2 = 8.
  2. Reactants: 16 + 2 × 18 = 52.
  3. 8 ÷ 52 × 100 = 15.4%.

Question 2 Medium

10.0 g of calcium carbonate is heated: CaCO₃ → CaO + CO₂. 3.5 g of calcium oxide is collected. Calculate the percentage yield to 3 significant figures. Mr of CaCO₃ = 100 and CaO = 56.

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Answer: 62.5%

  1. Moles of CaCO₃ = 10.0 ÷ 100 = 0.100.
  2. Theoretical yield of CaO = 0.100 × 56 = 5.6 g.
  3. 3.5 ÷ 5.6 × 100 = 62.5%.

Question 3 Medium

The titres in an experiment are 18.30, 18.35, 19.20 and 18.25 cm³. Which result is anomalous?

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Answer: 19.20 cm³

  1. The results should be close together.
  2. 19.20 cm³ is far from the others.

Question 4 Hard

N₂ + 3H₂ → 2NH₃. Calculate the mass of ammonia made from 6.0 g of hydrogen with excess nitrogen. Ar of H = 1 and N = 14.

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Answer: 34 g

  1. Moles of H₂ = 6.0 ÷ 2 = 3.0.
  2. The ratio H₂ : NH₃ is 3 : 2, so moles of NH₃ = 3.0 × 2 ÷ 3 = 2.0.
  3. Mr of NH₃ = 17, so mass = 2.0 × 17 = 34 g.

Question 5 Hard

Mg + 2HCl → MgCl₂ + H₂. 0.30 mol of magnesium is added to 0.50 mol of hydrochloric acid. Calculate the moles of hydrogen formed.

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Answer: 0.25 mol

  1. 0.30 mol of Mg needs 0.60 mol of HCl, but only 0.50 mol is present.
  2. So HCl is limiting.
  3. Moles of H₂ = 0.50 ÷ 2 = 0.25 mol.

Question 6 Hard

25.0 cm³ of hydrochloric acid of concentration 0.150 mol/dm³ is neutralised by sodium hydroxide solution of concentration 0.100 mol/dm³. HCl + NaOH → NaCl + H₂O. Calculate the volume of sodium hydroxide solution needed, in cm³.

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Answer: 37.5 cm³

  1. Moles of HCl = 0.150 × 0.0250 = 0.00375.
  2. The ratio is 1 : 1, so moles of NaOH = 0.00375.
  3. Volume = 0.00375 ÷ 0.100 = 0.0375 dm³ = 37.5 cm³.

Question 7 Hard

20.0 cm³ of sulfuric acid is neutralised by 30.0 cm³ of 0.100 mol/dm³ sodium hydroxide. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the acid in g/dm³. Mr of H₂SO₄ = 98.

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Answer: 7.35 g/dm³

  1. Moles of NaOH = 0.100 × 0.0300 = 0.00300.
  2. Moles of H₂SO₄ = 0.00300 ÷ 2 = 0.00150.
  3. Concentration = 0.00150 ÷ 0.0200 = 0.0750 mol/dm³, and 0.0750 × 98 = 7.35 g/dm³.

Question 8 Hard

Fe₂O₃ + 3CO → 2Fe + 3CO₂. 8.0 g of iron(III) oxide makes 4.2 g of iron. Calculate the percentage yield. Mr of Fe₂O₃ = 160 and Ar of Fe = 56.

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Answer: 75%

  1. Moles of Fe₂O₃ = 8.0 ÷ 160 = 0.050.
  2. Moles of Fe = 2 × 0.050 = 0.10, so theoretical mass = 0.10 × 56 = 5.6 g.
  3. Percentage yield = 4.2 ÷ 5.6 × 100 = 75%.

Question 9 Hard

A solution contains 0.050 mol of potassium hydroxide in 200 cm³. Calculate its concentration in g/dm³. Mr of KOH = 56.

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Answer: 14 g/dm³

  1. Concentration = 0.050 ÷ 0.200 = 0.25 mol/dm³.
  2. In g/dm³: 0.25 × 56 = 14.

Question 10 Hard

Calculate the atom economy for making aluminium by 2Al₂O₃ → 4Al + 3O₂ to 3 significant figures. Mr of Al₂O₃ = 102 and Ar of Al = 27.

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Answer: 52.9%

  1. Useful product: 4 × 27 = 108.
  2. Reactants: 2 × 102 = 204.
  3. 108 ÷ 204 × 100 = 52.9%.

Question 11 Hard

A student calculates a percentage yield of 108%. Suggest why this answer cannot be correct.

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Answer: a yield cannot be more than 100%, because the product may be wet or impure

  1. The actual yield cannot be bigger than the theoretical yield.
  2. The product is probably wet or impure, or there is a calculation error.

Quick checks

Short questions with options. Choose one before you open the answer.

Check 1

How do you change a mass in grams into moles?

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Divide by the Mr (or the Ar for an atom).

Check 2

How do you change moles into a mass?

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Multiply by the Mr.

Check 3

How do you change cm³ into dm³?

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Divide by 1000.

Check 4

How do you find moles of a solute from concentration and volume?

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Moles = concentration in mol/dm³ × volume in dm³.

Check 5

How do you change mol/dm³ into g/dm³?

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Multiply by the Mr.

Check 6

What are the four lines of a reacting mass calculation?

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Balanced equation, moles of the known substance, ratio, mass of the wanted substance.

Check 7

How do you find the limiting reactant?

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Work out the moles of each, use the equation ratio, and see which one runs out first.

Check 8

Which reactant do you use to find the amount of product?

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The limiting reactant.

Check 9

What is the equation for percentage yield?

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Actual yield ÷ theoretical yield × 100.

Check 10

What is the equation for atom economy?

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Mr of the useful product ÷ total Mr of the reactants × 100.

Check 11

Why is the rough titre left out of the mean?

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It was run quickly, so it is less accurate. Only concordant titres are used.

Check 12

What does concordant mean in a titration?

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Within 0.10 cm³ of each other.

Check 13

What are the five links in a titration calculation?

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Mean titre, moles of the known solution, ratio, concentration of the unknown, convert to g/dm³ if asked.

Check 14

In H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, how do the moles of acid and alkali compare?

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The acid moles are half the alkali moles.

Check 15

What is the highest possible percentage yield?

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100%. Anything more means an error or an impure product.

Check 16

Why can a reaction with a high yield still have a low atom economy?

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It can make a lot of waste by-products, which are not counted as useful product.

Check 17

What is the atom economy of a reaction with only one product?

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100%.

Check 18

What does the Avogadro constant equal?

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6.02 × 10²³ particles per mole.

Check 19

What does show that mean in a calculation question?

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Show every step and reach the value you have been given.

Check 20

How should you round a final answer?

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To the same significant figures as the least precise data, or as the question asks, and only at the end.

Check 21

What check shows a reacting mass answer is sensible?

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A product from one reactant cannot weigh more than the reactant, and product masses add up to reactant masses.

Check 22

How many moles of H₂ form from 0.50 mol of HCl in Mg + 2HCl → MgCl₂ + H₂?

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0.25 mol, because the ratio HCl : H₂ is 2 : 1.

Check 23

Why must volumes be in dm³ in a concentration calculation?

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Concentration is amount per dm³, so the volume must use the same unit.

Check 24

What does excess mean?

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More of a reactant than is needed, so some is left over.

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