Due to partial corrosion of a piece of copper into cuprous sulphide $C{u}_{2}S$, it gains weight. If the % of total copper that has undergone corrosion is 32%, then % gain in weight of piece of copper is: (Take atomic mass of Cu & S as 63 & 32 respectively).

Let mass of Cu piece = 100 g 
2Cu  +    $H_{2}S$   $\to$      $C{u}_{2}S$  +  $H_2$
32 g         –                     –                  –
$\frac{32}{32}=1mol$  –   –  –
–  –      $\frac{1}{2}mol$           $\frac{1}{2}mol$
                $(\frac{1}{2}\times 158)g$
Weight of Cu left + weight of  $C{u}_{2}S$ formed = 68 + 40.12= 108.12  g
Increase in weight due to partial corrosion of Cu = $8.12\approx 8$
$\therefore$ % gain in weight of piece of Cu = 47