What is the mass of Sulphur produced by 1 mole of ${\mathit{SO}}_2$ and 1 mole of $H_2\mathit{S }$which react completely to form 2 mole of $H_2\mathit{O }$and 3 moles of $S$?

${\mathit{SO}}_{\mathit{2}}$ +  $2H_2\mathit{S }$  $\to$  $2H_{2}O$  +  $3S$ 

(Atomic mass of S = 32, O = 16)

The given chemical reaction produces 48g of Sulphur.

1 mole of ${\mathit{SO}}_2$ and 2 moles of $H_{2}S$ react completely to form,
$\Rightarrow$  ${\mathit{SO}}_2$ +  $2H_2\mathit{S }$  $\to$  $2H_{2}O$  +  $3S$ 

When 2 mol of${\mathit{ H}}_{2}S$   react with 1 mole of  ${\mathit{SO}}_2$ , give 3 mol of $S$, but we are given 1 mol of $H_{2}S$and 1 mole of ${\mathit{SO}}_2$.

Hence, $H_{2}S$ is a limiting reagent, i.e. a reagent wholly consumed in a reaction which limits the amount of product that will be formed and hence determines the amount of product produced.
So, 1 mole of $H_{2}S$ will give $\frac{3}{2}$mol of $S$.

Hence,  mass of sulphur will be,
$\Rightarrow \left( \frac{3}{2}\mathit{mol}\right) \times (32g/\mathit{mol})$
$\Rightarrow 48g$

So, the mass of sulphur that is produced is 48 g.