3.7 Limiting reactants
5 exam questions worth 12 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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What is meant by the limiting reactant?
Choose one answer.
Correct Not quite: the answer is The reactant that is completely used up
Mark scheme
D
Model answer
D: the limiting reactant is completely used up, so it stops the reaction.
4.8 g of magnesium reacts with 0.40 mol of hydrochloric acid: Mg + 2HCl → MgCl2 + H2
Show that magnesium is the limiting reactant. (A_\mathrm{r}: Mg = 24)
Mark scheme
- moles Mg = 4.8 ÷ 24 = 0.20 (1)
- 0.20 mol Mg needs 0.40 mol HCl, so HCl is just enough / Mg is used up first (1)
Model answer
Moles of Mg = 0.20. This needs exactly 0.40 mol of HCl, so the magnesium is all used up; adding more acid would not make more product, so magnesium is limiting.
6.0 g of carbon is burned in 32 g of oxygen: C + O2 → CO2
Identify the limiting reactant and calculate the mass of carbon dioxide formed. (A_\mathrm{r}: C = 12, O = 16)
Mark scheme
- moles C = 0.50; moles O2 = 1.0 (1)
- carbon is limiting (1)
- 0.50 mol CO2 (1)
- 0.50 × 44 = 22 g (1)
Model answer
C = 0.50 mol and O2 = 1.0 mol. The ratio is 1 : 1, so carbon runs out first and is limiting. 0.50 mol of CO2 forms: 0.50 × 44 = 22 g.
Explain why the mass of product formed depends only on the mass of the limiting reactant.
Mark scheme
- the reaction stops when the limiting reactant is used up (1)
- the excess reactant has nothing left to react with (1)
Model answer
Once the limiting reactant has been used up, the reaction stops; the extra amount of the other reactant cannot react, so it does not make more product.
5.6 g of iron reacts with 3.2 g of sulfur to form 8.8 g of iron sulfide.
Use these masses to find the balanced equation. (A_\mathrm{r}: Fe = 56, S = 32)
Mark scheme
- moles Fe = 0.10, S = 0.10 (1)
- FeS: 8.8 ÷ 88 = 0.10 so ratio 1 : 1 : 1 (1)
- Fe + S → FeS (1)
Model answer
Fe = 0.10 mol, S = 0.10 mol and FeS (M_\mathrm{r} 88) = 0.10 mol, a 1 : 1 : 1 ratio: Fe + S → FeS.
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