Question

In the given reaction:
$H_{2}S\left(g\right)+S{O}_2\left(g\right)⟶S\left(s\right)+H_{2}O$
One equivalent of $H_{2}S\left(g\right)$ will reduce :

• A. $1$ mol of $S{O}_2$
• B. $0.5$ mol of $S{O}_2$
• C. $0.25$ mol of $S{O}_2$
• D. $2$ mol of $S{O}_2$

Answer 1

The correct option is C $0.25$ mol of $S{O}_2$

$2H_2\underset{\uparrow -2}{S}\left(g\right)+\underset{\uparrow +4}{S}{O}_2\left(g\right)⟶3\underset{0}{S}\left(s\right)+2H_{2}O\left(l\right)$
Change in oxidation number of each S in $H_{2}S=2$ units.
So, one mole of $H_{2}S=2$ equivalents.
According to balanced reaction, one mole of $H_{2}S$ (2 equivalent) reacts with 0.5 mole of $S{O}_2$.
So, one equivalent of $H_{2}S$ reacts with 0.25 mole of $S{O}_2$.