Question

Zn metal reduces ${S{O}_3}^{2-}$ ions into $H_{2}S$in presence of concentrated $H_{2}S{O}_4$ What weight of Zn is required for
reduction of 6.3 g $N{a}_{2}S{O}_3$ in presence of concentrated acid.

• A. 9.75 g
• B. 13 g
• C. 130 g
• D. 23 g

Answer 1

The correct option is A 9.75 g

eq. of $Zn=$ eq.of ${S{O}_3}^{2-}$
$\frac{w}{65}\times 2=\frac{6.3}{126}\times 6\Rightarrow w=\frac{6.3}{126}\times \frac{6\times 65}{2}=9.75g$