Reacting masses with a limiting reactant

Quantitative chemistry

GCSE Chemistry (8462) · 4.3.2.2 & 4.3.2.4 · Higher

ORIGINAL

Turn both masses into moles, use the equation's ratio to find which runs out, then work from that one.

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Question

Sodium burns in oxygen:

\mathrm{4Na(s) + O_2(g) \rightarrow 2Na_2O(s)}

. A student reacts 4.60 g of sodium with 1.00 g of oxygen. Identify the limiting reactant and calculate the maximum mass of sodium oxide formed.

A_r

: Na = 23, O = 16. Give your answer to 3 significant figures.

Every step worked, with the reasoning.

1
$n(\mathrm{Na}) = \dfrac{4.60}{23} = 0.200\ \text{mol}$
Convert each reactant's mass to moles.
2
$n(\mathrm{O_2}) = \dfrac{1.00}{32} = 0.03125\ \text{mol}$
Oxygen is diatomic, so use $M_r(\mathrm{O_2}) = 32$ , not 16.
3
$0.200\ \text{mol Na needs } \dfrac{0.200}{4} = 0.0500\ \text{mol } \mathrm{O_2}$ , but only $0.03125\ \text{mol}$ is present, so oxygen is the limiting reactant
Use the 4 : 1 ratio from the equation. Common error: deciding from the masses (4.60 g against 1.00 g) or ignoring the ratio — always compare moles through the equation.
4
$n(\mathrm{Na_2O}) = 2 \times 0.03125 = 0.0625\ \text{mol}$
Work from the limiting reactant: 1 mol of O₂ makes 2 mol of Na₂O.
5
$M_r(\mathrm{Na_2O}) = 2(23) + 16 = 62$ , so $\text{mass} = 0.0625 \times 62 = 3.875\ \text{g}$
Mass = moles × $M_r$ . Check by conservation of mass: the 0.125 mol of sodium that reacts is 2.875 g, and 2.875 + 1.00 = 3.875 g.

Answer: Oxygen is limiting; maximum mass $= 3.88\ \text{g}$ (3 s.f.)