Question

2 moles of $H_{2}S$ and 11.2 L of $S{O}_2$ at NTP react to give x mole of sulphur. The Value of 'x' will be:
Reaction: $S{O}_2\left(g\right)+2H_{2}S\left(g\right)\to 3S+2H_{2}O$

• A. 1.5
• B. 3
• C. 11.2
• D. 6

Answer 1

The correct option is A 1.5

we have, 11.2 L of $S{O}_2$ = 0.5 mole

for 1 mole of $S{O}_2$ we require 2 moles of $H_{2}S$

and for 0.5 mole of $S{O}_2$ we require 1 moles of $H_{2}S$

so, we have $S{O}_2$ as a limiting reagent

1 mole of $S{O}_2$ form 3 mole of sulphur

0.5 mole of $S{O}_2$ produce $S$ = 0.5 * 3 = 1.5 mole