Volume of gas at room temperature and pressure

Quantitative chemistry

GCSE Chemistry (8462) · 4.3.5 · Higher

ORIGINAL

Moles of gas × 24 dm³ at RTP — then convert to the unit the question asks for (chemistry only).

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Question

Calcium carbonate reacts with excess hydrochloric acid:

\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)}

. Calculate the volume of carbon dioxide, in cm³, produced at room temperature and pressure (RTP) from 5.00 g of calcium carbonate.

A_r

: Ca = 40, C = 12, O = 16. One mole of any gas occupies 24 dm³ at RTP.

Every step worked, with the reasoning.

1
$M_r(\mathrm{CaCO_3}) = 40 + 12 + 3(16) = 100$ , so $n(\mathrm{CaCO_3}) = \dfrac{5.00}{100} = 0.0500\ \text{mol}$
The acid is in excess, so the calcium carbonate decides how much gas forms.
2
$n(\mathrm{CO_2}) = 0.0500\ \text{mol}$
The equation's ratio of calcium carbonate to carbon dioxide is 1 : 1.
3
$V = 0.0500 \times 24 = 1.20\ \text{dm}^3$
Volume of gas at RTP = moles × 24 dm³. Common error: mixing units — 24 gives dm³ and 24 000 gives cm³, so an answer of 1.20 labelled cm³ (or 1200 labelled dm³) loses the mark.
4
$1.20\ \text{dm}^3 \times 1000 = 1200\ \text{cm}^3$
The question asks for cm³, and 1 dm³ = 1000 cm³.

Answer: $1200\ \text{cm}^3$ , i.e. $1.20 \times 10^{3}\ \text{cm}^3$ (3 s.f.)