Question

In the following reaction,

$2H_{2}S+S{O}_2\to 2H_{2}O+3S$

Calculate the amount of sulphur when 1 mole of $S{O}_2$ reacts with 1 mole of $H_{2}S$.

• A. 1 mole
• B. 3 moles
• C. 1.5 moles
• D. 2 moles

Answer 1

The correct option is C 1.5 moles

$2H_{2}S+S{O}_2\to 2H_{2}O+3S$
2 moles of $H_{2}S$ reacts with 1 mole of $S{O}_2$ to give 3 moles of S.
1 mole of $H_{2}S$ and 1 mole of $S{O}_2$ are present.
$H_{2}S$ is limiting reagent and $S{O}_2$ is excess reagent.
1 mole of $H_{2}S$ will react with 0.5 mole of $S{O}_2$ to give $3\times \frac{1}{2}=1.5$ moles of S.