Question

In the following reaction:
$S{O}_2\left(g\right)+2H_{2}S\left(g\right)\to 3S\left(s\right)+2H_{2}O\left(l\right)$, the number of moles sulphur formed by $2$ moles each of $S{O}_2$ and $H_{2}S$ is :

Answer 1

$S{O}_2\left(g\right)+2H_{2}S\left(g\right)\to 3S\left(s\right)+2H_{2}O\left(l\right)$
As per equation $1$ mol $2$ mol $3$ mol $2$ mol
taken : $2$ mol $2$ mol $?$
If one takes $2$ moles $S{O}_2$, then $4$ moles of $H_{2}S$ are required.
But $2$ moles of $H_{2}S$ have been taken.
Hence, $H_{2}S$ is limiting reagent that forms $3$ moles of sulphur.
Thus, $3$ moles of sulphur is formed by $2$ moles each of $S{O}_2$ and $H_{2}S$.