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  4. 1 mol of SO2 and 1 mol of H2S react completely to form H2O and S as follows :SO2 + 2H2S arrow 2H2O + 3 S (At. mass S = 32,O = 16)The mass of S obtained is;

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Q. 1 mol of $SO_2$ and 1 mol of $H_2S$ react completely to form $H_2O$ and S as follows :$SO_2 + 2H_2S \rightarrow 2H_2O + 3 S$ (At. mass S = 32,O = 16)The mass of S obtained is;

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Solution:

$SO_{2}+H_{2}S \rightarrow2H_{2}O+3S$
Here $H_{2}S$ is the limiting reactant
$2$ mol of $H_{2}S$ give $3$ mol of $S$
$\therefore $ $1$ mol of $H_{2}S$ gives $3/2$ mol of $S$
Mass of sulphur $=\left(\frac{3}{2}mol\right)\times\left(32\,g\,mol^{-1}\right)$
$=48\,g$