Question

At STP, $2.8$ litres of hydrogen sulphide were mixed with $1.6$ litre of sulphur dioxide and the reaction occurred according to the equation:
$2H_{2}S\left(g\right)+S{O}_2\left(g\right)\to 2H_{2}O\left(l\right)+3S\left(s\right)$
Which of the following shows the volume of the gas remaining after the reaction?

• A. $0.2$ litres of $S{O}_2\left(g\right)$
• B. $0.4$ litres of $H_2\left(g\right)$
• C. $1.2$ litres of $H_{2}S\left(s\right)$
• D. $1.2$ litres of $S{O}_2\left(g\right)$

Answer 1

The correct option is A $0.2$ litres of $S{O}_2\left(g\right)$

$2H_{2}S\left(g\right)+S{O}_2\left(g\right)\to 2H_{2}O\left(l\right)+3S\left(s\right)$

$21$

Therefore,

2 moles of $H_{2}S$ is needed for 1 mole of $S{O}_2$

Thus, for 1.6 litres of $S{O}_2$ we require about 3.2 litres of $H_{2}S$

$\therefore 2.8H_{2}S$ is limiting reagent

Thus, for 2.8 litres of $H_{2}S$ only 1.4 lites of $S{O}_2$ will be used.

So remaining $=1.6-1.4=0.2ltrofS{O}_2$