Question

What amount of Zn will be required to produce $H_2$ by its action on dilute $H_2{\mathrm{SO}}_4$ which will completely react with the oxygen, produced by heating 30g of ${\mathrm{KClO}}_3$?

Answer 1

${\mathrm{KCl}}_3\to \mathrm{KCl}+\frac{3}{2}{O}_2$
Amount 30 g
122.5 g of ${\mathrm{KClO}}_3$will gives = 48 g of oxygen
30 g of ${\mathrm{KClO}}_3$will give = $\frac{48}{122.5}\times 30$ = 11.755 g
$H_2+\frac{1}{2}{O}_2\to H_{2}O$
16g of oxygen reacts with = 2g of $H_2$
11.755g of oxygen will react with = $\frac{2}{16}\times 11.755$ $=1.469g\mathrm{of}{H}_2$
$\mathrm{Zn}+H_2{\mathrm{SO}}_4\to {\mathrm{ZnSO}}_4+H_2$
2g of $H_2$ is yielded from = 63.5g of Zn
1.469g of $H_2$ will be yield from= $\frac{65.3}{2}\times 1.469$ = 47.976g of Zn.