These questions check the reasoning from the three lessons in conservation reasoning across a reaction. Use Ar values: H = 1, C = 12, N = 14, O = 16, Mg = 24, Al = 27, Fe = 56, Cu = 64, Ag = 108.
Work each question on paper first, then open the answer. Molar volume is 24 dm³ mol⁻¹ at room conditions.
Questions
Question 1
N₂ + 3H₂ → 2NH₃. How many moles of ammonia form from 0.30 mol of nitrogen?
Answer
The ratio N₂ : NH₃ is 1 : 2, so moles of NH₃ = 0.30 × 2 = 0.60 mol. The coefficient 3 in front of H₂ is not needed here.
Question 2
2Mg + O₂ → 2MgO. A strip of magnesium of mass 4.8 g burns completely. Find the mass of magnesium oxide and the mass of oxygen used.
Answer
Moles of Mg = 4.8 ÷ 24 = 0.20 mol. Moles of MgO = 0.20 mol, so mass = 0.20 × 40 = 8.0 g.
Moles of O₂ = 0.20 ÷ 2 = 0.10 mol, so mass = 0.10 × 32 = 3.2 g. Check: 4.8 + 3.2 = 8.0 g, so mass is conserved.
Question 3
A student says that in 2H₂O the coefficient 2 means 2 g of water. Explain the error, then find the mass of water formed from 0.50 mol of hydrogen in 2H₂ + O₂ → 2H₂O.
Answer
A coefficient counts particles or moles, not grams. Moles of H₂O = 0.50 mol because the ratio is 2 : 2. Mass = 0.50 × 18 = 9.0 g.
Question 4
2Al + 6HCl → 2AlCl₃ + 3H₂. 2.7 g of aluminium reacts with 0.20 mol of hydrochloric acid. Find the limiting reactant and the volume of hydrogen formed.
Answer
Moles of Al = 2.7 ÷ 27 = 0.10 mol. Divide by coefficients: Al gives 0.10 ÷ 2 = 0.050, and HCl gives 0.20 ÷ 6 = 0.033.
The smaller value belongs to HCl, so HCl is limiting. Moles of H₂ = 0.20 × 3 ÷ 6 = 0.10 mol, so volume = 0.10 × 24 = 2.4 dm³.
Question 5
6.20 g of copper(II) carbonate is heated: CuCO₃ → CuO + CO₂. The black solid weighs 3.40 g. Find the percentage yield and give one reason it is below 100%.
Answer
Mr of CuCO₃ = 124, so moles = 6.20 ÷ 124 = 0.0500 mol. Calculated CuO = 0.0500 × 80 = 4.00 g.
Percentage yield = 3.40 ÷ 4.00 × 100 = 85.0%. A valid reason: some carbonate did not fully decompose, or some solid was lost on transfer.
Question 6
A student writes Fe + Ag⁺ → Fe²⁺ + Ag. Check the equation for mass and for charge separately, then correct it.
Answer
Mass check: one Fe and one Ag on each side, so atoms balance. Charge check: left side +1, right side +2, so charge does not balance.
Correct equation: Fe + 2Ag⁺ → Fe²⁺ + 2Ag. Now charge is +2 on both sides and there are two Ag atoms on each side.
Question 7
A student reports a percentage yield of 105%. State one likely reason, and explain why it cannot mean that extra mass was created.
Answer
The product was probably not fully dry, or it contained an impurity, so the measured mass was too large. Matter cannot be created, so the extra mass must come from something else that was weighed with the product, such as water or leftover reactant.
If you got these wrong
- Questions 1 to 3 went wrong: revisit connecting particle ratios to mole ratios.
- Question 4 went wrong: read identifying the limiting reactant.
- Questions 5 and 7 went wrong: read explaining a smaller measured yield.
- Question 6 went wrong: read checking mass and charge independently in an ionic equation.
Keep a record of which step fails each time with the mistake log and paper error review. If the same step keeps failing, see online one-to-one Chemistry tuition.