Limiting reactants questions and answers

Worked solution

1. Mg needs twice as many moles of HCl: $0.2 \times 2 = 0.4$ mol
2. Only 0.3 mol HCl is there, so HCl runs out first.

Worked solution

1. $\text{H}_2$ needs half as much $\text{O}_2$: $3 \div 2 = 1.5$ mol, and there's 2 mol, so $\text{H}_2$ is limiting.
2. 2 H₂ : 2 H₂O, so 3 mol water.

Worked solution

1. $2 - 1.5 = 0.5$ mol of $\text{O}_2$ in excess

Worked solution

1. CuO: $8 \div 79.5 = 0.1006$ mol
2. $\text{H}_2\text{SO}_4$: $9.8 \div 98 = 0.1$ mol
3. Ratio 1 : 1, so $\text{H}_2\text{SO}_4$ (fewer moles) is limiting.

Worked solution

1. Na: $4.6 \div 23 = 0.2$ mol; $\text{Cl}_2$: $7.1 \div 71 = 0.1$ mol
2. 0.2 mol Na needs exactly 0.1 mol $\text{Cl}_2$: neither is in excess.
3. 0.2 mol NaCl $\times 58.5 = 11.7$ g

Worked solution

1. Zn: $13 \div 65 = 0.2$ mol, needing 0.4 mol HCl; only 0.2 mol is there, so HCl is limiting.
2. 2 HCl gives 1 $\text{H}_2$: $0.2 \div 2 = 0.1$ mol

Worked solution

1. $\text{N}_2$: $28 \div 28 = 1$ mol; $\text{H}_2$: $4 \div 2 = 2$ mol
2. 1 mol $\text{N}_2$ needs 3 mol $\text{H}_2$, so hydrogen is limiting.
3. $\text{NH}_3 = 2 \times \tfrac{2}{3} = 1.333$ mol; mass $= 1.333 \times 17 = 22.67$ g

Worked solution

1. Using an excess of one reactant makes sure all of the other reactant is used, so none of it is wasted.