The questions are on the reacting masses test. Take it under real conditions before reading the mark scheme.
Quick answer: Reacting masses come from a balanced equation. This mark scheme gives the model answers and the marks for the 9 questions of the Chemistry 9-10 test on reacting masses, from balance and ratio to silver nitrate solution, with the equivalent answers to accept.
Model answers and mark schemes for the test. Award marks for correct method even when the final value is wrong, and apply error carried forward where the question allows.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-7, GCSE C3
Question 1 Balance and ratio (2 marks)
- 2Fe + 3Cl₂ → 2FeCl₃.
- n(Cl₂) = 0.40 × 3/2 = 0.60 mol.
Marking: 1 mark for balanced equation; 1 mark for the 3 : 2 ratio used correctly.
Question 2 Salt from sodium carbonate (3 marks)
- M = 106; n = 5.30 ÷ 106 = 0.0500 mol.
- Ratio 1 : 2, so n(NaCl) = 0.100 mol.
- m = 0.100 × 58.5 = 5.85 g.
Marking: 1 mark each; allow error carried forward.
Question 3 Phosphorus and oxygen (3 marks)
- 0.20 mol of P needs 0.20 × 5/4 = 0.25 mol of O₂. There is 0.30 mol of O₂, which is more than 0.25 mol, so oxygen is in excess and phosphorus is limiting.
- n(P₂O₅) = 0.20 × 2/4 = 0.10 mol; M = 142; m = 14.2 g.
Marking: 1 mark for moles of oxygen needed; 1 mark for conclusion; 1 mark for mass.
Question 4 Yield and losses (2 marks)
- 5.6 ÷ 8.0 × 100 = 70%.
- Some product is lost when it is transferred or filtered, or some reactant forms by-products.
Marking: 1 mark for the percentage; 1 mark for any valid reason.
Question 5 Concentration of a solution (2 marks)
- n = 2.00 ÷ 40 = 0.0500 mol; c = 0.0500 ÷ 0.500 = 0.100 mol/dm³.
- n = 0.100 × 0.0250 = 0.00250 mol.
Marking: 1 mark for each correct value, with the volume converted to dm³.
Question 6 Mass concentration (2 marks)
- 2.00 ÷ 0.500 = 4.00 g/dm³.
- Each dm³ holds twice as many moles, so there are twice as many solute particles in the same volume.
Marking: 1 mark each.
Question 7 Reading a graph (2 marks)
- The line levels off at 0.6 g.
- All the acid has reacted, so the extra magnesium is in excess and no more product can form.
Marking: 1 mark each.
Question 8 Explain the limiting reactant (2 marks)
- The fuel is limiting.
- The fuel has all been used up, while the oxygen in the air is in excess, so burning cannot continue.
Marking: 1 mark each.
Question 9 Silver nitrate solution (2 marks)
- n = 0.050 × 0.100 = 0.0050 mol.
- M = 170; m = 0.0050 × 170 = 0.85 g.
Marking: 1 mark each; allow error carried forward.
