2019 H2 Chem P1 Q7 — Mole Solution

The answer is D (\(2.95 \text{ dm}^3\)).

\(2\text{NaN}_3\text{(s)} \rightarrow 2\text{Na(s)} + 3\text{N}_2\text{(g)}\)
\(M_r\) of \(\text{NaN}_3 = 23.0 + 3(14.0) = 65.0\)
Amount of \(\text{NaN}_3 = \dfrac{5.00}{65.0} = 0.07692\) mol
Amount of \(\text{N}_2 = \dfrac{3}{2} \times 0.07692 = 0.1154\) mol
T = 273 + 30.0 = 303 K
\(V = \dfrac{nRT}{p} = \dfrac{0.1154 \times 8.31 \times 303}{9.85 \times 10^4} = 2.95 \times 10^{-3} \text{ m}^3 = 2.95 \text{ dm}^3\)

• A is wrong: \(0.00295\) is the volume in \(\text{m}^3\); it has not been converted to \(\text{dm}^3\) (1 \(\text{m}^3\) = 1000 \(\text{dm}^3\)).
• B is wrong: \(0.292 \text{ dm}^3\) comes from using T = 30.0 instead of 303 K; the ideal gas equation needs temperature in kelvin.
• C is wrong: \(1.97 \text{ dm}^3\) comes from taking the amount of \(\text{N}_2\) as 0.07692 mol, the amount of \(\text{NaN}_3\), and missing the 2 : 3 mole ratio in the equation.