These questions revise the lessons in moles, formulas and equations. Use Ar values: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, Al = 27, S = 32, Ca = 40, P = 31.
Try each question on paper before you open the answer. Use 6.02 × 10²³ mol⁻¹ and 24 dm³ mol⁻¹ at room conditions.
Questions
Question 1
Find the mass of 0.30 mol of sodium carbonate, Na₂CO₃.
Answer
Molar mass = 2 × 23 + 12 + 3 × 16 = 106 g mol⁻¹. Mass = 0.30 × 106 = 31.8 g.
Question 2
How many atoms are there in 3.2 g of sulfur?
Answer
Moles = 3.2 ÷ 32 = 0.10 mol. Sulfur is counted as atoms here, so atoms = 0.10 × 6.02 × 10²³ = 6.02 × 10²².
Question 3
A compound contains 3.10 g of phosphorus and 4.00 g of oxygen. Find its empirical formula.
Answer
Moles of P = 3.10 ÷ 31 = 0.100. Moles of O = 4.00 ÷ 16 = 0.250. Dividing by 0.100 gives P 1 and O 2.5.
The ratio 1 : 2.5 needs doubling, giving 2 : 5. The empirical formula is P₂O₅.
Question 4
Balance Al + O₂ → Al₂O₃, then find the mass of aluminium oxide formed from 5.4 g of aluminium.
Answer
Balanced equation: 4Al + 3O₂ → 2Al₂O₃.
Moles of Al = 5.4 ÷ 27 = 0.20 mol. The ratio Al : Al₂O₃ is 4 : 2, so moles of Al₂O₃ = 0.10 mol. Mass = 0.10 × 102 = 10.2 g.
Question 5
10.0 g of calcium carbonate reacts with excess hydrochloric acid. Find the volume of carbon dioxide at room conditions.
Answer
Moles of CaCO₃ = 10.0 ÷ 100 = 0.100 mol. The ratio CaCO₃ : CO₂ is 1 : 1, so moles of CO₂ = 0.100 mol. Volume = 0.100 × 24 = 2.4 dm³.
Question 6
4.0 g of sodium hydroxide is dissolved to make 250 cm³ of solution. Find the concentration in mol dm⁻³.
Answer
Moles of NaOH = 4.0 ÷ 40 = 0.10 mol. Volume = 250 ÷ 1000 = 0.250 dm³. Concentration = 0.10 ÷ 0.250 = 0.40 mol dm⁻³.
Question 7
25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is neutralised by 0.0500 mol dm⁻³ sulfuric acid: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Find the volume of acid used.
Answer
Moles of NaOH = 0.100 × 25.0 ÷ 1000 = 0.00250 mol. The ratio NaOH : H₂SO₄ is 2 : 1, so moles of acid = 0.00125 mol.
Volume = 0.00125 ÷ 0.0500 = 0.0250 dm³ = 25.0 cm³.
Question 8
2Mg + O₂ → 2MgO. 2.4 g of magnesium is heated with 0.15 mol of oxygen. Find the limiting reactant and the mass of magnesium oxide.
Answer
Moles of Mg = 2.4 ÷ 24 = 0.10 mol. Divide by coefficients: Mg gives 0.10 ÷ 2 = 0.050, and O₂ gives 0.15 ÷ 1 = 0.15.
The smaller value belongs to Mg, so magnesium is limiting. Moles of MgO = 0.10 mol, so mass = 0.10 × 40 = 4.0 g.
If you got these wrong
- Questions 1 and 2: revise converting between mass, moles and particles.
- Question 3: revise finding empirical and molecular formulas.
- Questions 4 to 7: revise reacting-mass and gas-volume calculations.
- Question 8: revise identifying the limiting quantity.
Record which step failed with the mistake log and paper error review, and plan a timed set with the timed original practice session builder. If one step keeps failing, see online one-to-one Chemistry tuition.