AQA GCSE Combined Science: Trilogy · 5.3

Quantitative chemistry

50 revision questions with answers for 5.3 Quantitative chemistry in AQA GCSE Combined Science: Trilogy (spec 8464). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below.

All 50 questions and answers

Question 1 · Spec 5.3.1.1
State the law of conservation of mass.
Answer:
No atoms are created or destroyed in a chemical reaction, so total reactant mass equals total product mass.
Question 2 · Spec 5.3.1.1
What must be equal on both sides of a balanced equation?
Answer:
The number of atoms of each element.
Question 3 · Spec 5.3.1.1
What is the difference between 2H₂O and H₂O₂?
Answer:
2H₂O means two water molecules. H₂O₂ is one molecule of a different compound, hydrogen peroxide.
Question 4 · Spec 5.3.1.1
Why must coefficients rather than formula subscripts be changed when balancing?
Answer:
Changing a subscript changes the substance. Coefficients change only the numbers of particles.
Question 5 · Spec 5.3.1.1
Balance Mg + O₂ → MgO.
Answer:
2Mg + O₂ → 2MgO.
Question 6 · Spec 5.3.1.1
Balance H₂ + O₂ → H₂O.
Answer:
2H₂ + O₂ → 2H₂O.
Question 7 · Spec 5.3.1.1
Balance CH₄ + O₂ → CO₂ + H₂O.
Answer:
CH₄ + 2O₂ → CO₂ + 2H₂O.
Question 8 · Spec 5.3.1.1
In a closed reaction, 12 g of A reacts with 8 g of B. Find total product mass.
Answer:
Total product mass = 12 + 8 = 20 g.
Question 9 · Spec 5.3.1.2
How is relative formula mass, Mr, calculated?
Answer:
Add the relative atomic masses of all atoms shown in the formula, including atoms inside multiplied brackets.
Question 10 · Spec 5.3.1.2
Calculate Mr of H₂O. Ar: H = 1, O = 16.
Answer:
Mr = 2 × 1 + 16 = 18.
Question 11 · Spec 5.3.1.2
Calculate Mr of Ca(OH)₂. Ar: Ca = 40, O = 16, H = 1.
Answer:
Mr = 40 + 2 × (16 + 1) = 74.
Question 12 · Spec 5.3.1.2
Calculate Mr of Al₂(SO₄)₃. Ar: Al = 27, S = 32, O = 16.
Answer:
Mr = 2 × 27 + 3 × (32 + 4 × 16) = 342.
Question 13 · Spec 5.3.1.2
Does relative formula mass have a unit?
Answer:
No. It is a relative quantity, calculated from relative atomic masses.
Question 14 · Spec 5.3.1.2
Check mass balance for 2Mg + O₂ → 2MgO. Ar: Mg = 24, O = 16.
Answer:
Reactants: 2 × 24 + 32 = 80. Products: 2 × 40 = 80. The totals agree.
Question 15 · Spec 5.3.1.3
Why does magnesium gain mass when burned in air?
Answer:
Oxygen atoms from the air combine with magnesium; the oxygen's mass becomes part of the solid oxide.
Question 16 · Spec 5.3.1.3
Why can a carbonate lose measured mass during heating?
Answer:
Thermal decomposition releases carbon dioxide, which escapes from an open container.
Question 17 · Spec 5.3.1.3
Why would the total mass remain constant if all gases were retained?
Answer:
The same atoms remain within the closed system, even though they rearrange into new substances.
Question 18 · Spec 5.3.1.4
What is measurement uncertainty?
Answer:
The interval within which the true value can reasonably be expected to lie, based on measurement limitations and variation.
Question 19 · Spec 5.3.1.4
Calculate the mean of 10.2, 10.4 and 10.3 cm³.
Answer:
Mean = (10.2 + 10.4 + 10.3) / 3 = 10.3 cm³.
Question 20 · Spec 5.3.1.4
What is the range of measurements 10.2, 10.4 and 10.3 cm³?
Answer:
Range = maximum - minimum = 10.4 − 10.2 = 0.2 cm³.
Question 21 · Spec 5.3.1.4
Using half the range, estimate uncertainty for values from 9.8 to 10.4 g.
Answer:
Half-range uncertainty = (10.4 − 9.8) / 2 = ±0.3 g.
Question 22 · Spec 5.3.1.4
What should you do with a result far from the other repeats?
Answer:
Investigate a possible cause and repeat if appropriate. Exclude it only with a justified reason, rather than automatically deleting it.
Question 23 · Spec 5.3.1.4
Round 0.026784 to three significant figures.
Answer:
0.0268. Leading zeros do not count as significant figures.
Question 24 · Spec 5.3.2.5
Convert 250 cm³ to dm³.
Answer:
250 / 1,000 = 0.250 dm³.
Question 25 · Spec 5.3.2.5
What does a concentration of 20 g/dm³ mean?
Answer:
There are 20 g of dissolved solute in each 1 dm³ of solution.
Question 26 · Spec 5.3.2.5
What mass of solute is in 0.50 dm³ of a 12 g/dm³ solution?
Answer:
Mass = concentration x volume = 12 × 0.50 = 6.0 g.
Question 27 · Spec 5.3.2.5
Find solute mass in 200 cm³ of a 15 g/dm³ solution.
Answer:
200 cm³ = 0.200 dm³; mass = 15 × 0.200 = 3.0 g.
Question 28 · Spec 5.3.2.5
What volume of an 8.0 g/dm³ solution contains 2.0 g solute?
Answer:
Volume = mass / concentration = 2.0 / 8.0 = 0.25 dm³.
Question 29 · Spec 5.3.2.5
Find concentration when 5.0 g solute makes 0.20 dm³ of solution.
Answer:
Concentration = mass / volume = 5.0 / 0.20 = 25 g/dm³.
Question 30 · Spec 5.3.2.5 · Higher only
How does doubling solute mass at fixed solution volume affect concentration?
Answer:
Concentration doubles, because concentration is directly proportional to solute mass when volume is constant.
Question 31 · Spec 5.3.2.5 · Higher only
A solution is diluted from 100 to 500 cm³ with no solute lost. How does concentration change?
Answer:
The same solute occupies five times the volume, so concentration becomes one-fifth of the original.
Question 32 · Spec 5.3.2.1 · Higher only
What is a mole?
Answer:
An amount containing 6.02 × 10²³ stated particles, such as atoms, molecules or ions.
Question 33 · Spec 5.3.2.1 · Higher only
How is one mole's mass related to Mr?
Answer:
Its mass in grams is numerically equal to Mr; for example, one mole of CO₂ has mass 44 g.
Question 34 · Spec 5.3.2.1 · Higher only
Calculate moles in 12 g of magnesium. Ar(Mg) = 24.
Answer:
Amount = mass / molar mass = 12 / 24 = 0.50 mol.
Question 35 · Spec 5.3.2.1 · Higher only
Calculate the mass of 0.25 mol of CO₂. Mr = 44.
Answer:
Mass = amount x molar mass = 0.25 × 44 = 11 g.
Question 36 · Spec 5.3.2.1 · Higher only
How many molecules are in 2.0 mol of water?
Answer:
2.0 × 6.02 × 10²³ = 1.204 × 10²⁴ molecules, about 1.2 × 10²⁴ to two significant figures.
Question 37 · Spec 5.3.2.1 · Higher only
How many moles contain 3.01 × 10²³ atoms?
Answer:
Amount = number / Avogadro constant = 0.50 mol.
Question 38 · Spec 5.3.2.1 · Higher only
How many moles of oxygen atoms are in 1 mol of O₂ molecules?
Answer:
2 mol of oxygen atoms, because every O₂ molecule contains two oxygen atoms.
Question 39 · Spec 5.3.2.2 · Higher only
What mole ratio does Mg + 2HCl → MgCl₂ + H₂ show?
Answer:
1 mol Mg reacts with 2 mol HCl to form 1 mol MgCl₂ and 1 mol H₂.
Question 40 · Spec 5.3.2.2 · Higher only
For 2H₂ + O₂ → 2H₂O, how much water forms from 3 mol O₂ with excess H₂?
Answer:
The O₂:H₂O ratio is 1:2, so 6 mol H₂O forms.
Question 41 · Spec 5.3.2.2 · Higher only
For 2Mg + O₂ → 2MgO, find MgO mass from 6.0 g Mg. Ar: Mg = 24, O = 16.
Answer:
Mg amount = 6.0 / 24 = 0.25 mol. MgO amount = 0.25 mol; mass = 0.25 × 40 = 10 g.
Question 42 · Spec 5.3.2.2 · Higher only
CaCO₃ → CaO + CO₂. Find CO₂ mass from 25 g CaCO₃. Mr: CaCO₃ = 100, CO₂ = 44.
Answer:
CaCO₃ amount = 0.25 mol; CO₂ amount = 0.25 mol; mass = 0.25 × 44 = 11 g.
Question 43 · Spec 5.3.2.2 · Higher only
2H₂ + O₂ → 2H₂O. What mass of O₂ reacts with 4 g H₂? Mr: H₂ = 2, O₂ = 32.
Answer:
4 / 2 = 2 mol H₂ needs 1 mol O₂, which has mass 32 g.
Question 44 · Spec 5.3.2.3 · Higher only
24 g Mg reacts with 16 g O₂ to form 40 g MgO. Derive the equation. Mr: Mg = 24, O₂ = 32, MgO = 40.
Answer:
Amounts are 1, 0.5 and 1 mol. Multiply by two: 2Mg + O₂ → 2MgO.
Question 45 · Spec 5.3.2.3 · Higher only
4 g H₂ reacts with 32 g O₂ to form 36 g H₂O. Find coefficients. Mr: 2, 32 and 18 respectively.
Answer:
Amounts are 2, 1 and 2 mol, giving 2H₂ + O₂ → 2H₂O.
Question 46 · Spec 5.3.2.4 · Higher only
What is a limiting reactant?
Answer:
The reactant completely used up first, determining the maximum amount of product that can form.
Question 47 · Spec 5.3.2.4 · Higher only
Why might a chemist use one reactant in excess?
Answer:
To ensure the other reactant is completely used up.
Question 48 · Spec 5.3.2.4 · Higher only
2H₂ + O₂ → 2H₂O. Which limits a mixture of 3 mol H₂ and 1 mol O₂?
Answer:
O₂ limits the reaction: 1 mol O₂ needs only 2 mol H₂, leaving 1 mol H₂ unused.
Question 49 · Spec 5.3.2.4 · Higher only
2Mg + O₂ → 2MgO. Find product mass from 12 g Mg and excess oxygen. Mr: Mg = 24, MgO = 40.
Answer:
Mg limits the product: 12 / 24 = 0.50 mol; MgO mass = 0.50 × 40 = 20 g.
Question 50 · Spec 5.3.2.2 · Higher only
What sequence helps avoid errors in reacting-mass calculations?
Answer:
Balance the equation, convert known mass to moles, apply the mole ratio, then convert required moles to mass.

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