At STP, how many litres of hydrogen can be obtained from the reaction of \(500 cm ^{3}\) of \(0.5 M H _{2} SO _{4}\) with excess zinc metal
A. \(22.4 dm ^{3}\) B. \(11.2 dm ^{3}\) C. \(65 dm ^{3}\) D. \(5.6 dm ^{3} \) E. \(0.00 dm ^{3}\)
Correct Answer: D
Explanation
\begin{aligned} &\text { }\\ &Zn _{( s )}+ H _{2} SO _{4(\text { (aq) }} \rightarrow ZnSO _{4(\text { (aq) }}+ H _{2(\text { (g) }}\\ &\eta_{ H _{2} SO _{4}}: \eta_{ H _{2}}=1: 1\\ &\text { i.e. }\\ &\eta_{H_{3} SO _{4}}=\eta_{I_{2}}\\ &\eta_{ H _{2} SO _{4}}=C_{ H _{2} sO _{4}} \times V_{ H _{2} sO }\\ &=0.5 M \times 500 \times 10^{-3} dm ^{3}\\ &=0.25 mol\\ &\eta_{ HI _{2}}=\eta_{ H _{2} SO }=0.25 mol\\ &\text { No of mole }=\frac{\text { Volume at S.T.P }}{\text { Molar volume }}\\ &0.25 mol =\frac{\text { Volume of } H _{2} \text { at S.T.P }}{22.4 dm ^{3} / mol }\\ &\text { Vol. of } H _{2} \text { at s.t.p }\\ &=0.25 mol \times 22.4 dm ^{3} / mol\\ &=5.625 dm ^{3} \text { (1d.p) } \end{aligned}
