Question

What weight of zinc will react with dilute sulphuric acid to liberate 1000 mL of hydrogen at ${27}^{o}C$ and 750 mm of $Hg$ pressure?

Answer 1

$Zn+H_{2}S{O}_4⟶ZnS{O}_4+H_2$

From ideal equation, $PV=nRT.$

$n=\frac{PV}{RT}⟹\frac{\frac{750}{760}\times 1}{0.082\times 300}n=0.039$

Weight of $H_2=0.039\times 2=0.078g.$

$\because 2g$ of $H_2$ produced from $65g$ of $Zn.$

$\therefore 0.078g$ of $H_2$ produces from $\frac{65\times 0.078}{2}=2.53g$ of $Zn.$