Question

At NTP, $10$ liter of hydrogen sulphide gas reacted with $10$ liter of suphur dioxide gas. The volume of gas, after the reaction is completed would be:

• A. $5$ litre
• B. $10$ litre
• C. $15$ litre
• D. $20$ litre

Answer 1

Answer: (A)

$5$ litre

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

At NTP, $22.4L$ of $H_{2}S$ gas = $1$ mole

$10L$ of $H_{2}S$ gas = $\frac{1}{22.4}\times 10=0.446$ moles

At NTP, $22.4L$ of $S{O}_2$ gas = $1$ mole

$10L$ of $S{O}_2$ gas = $\frac{1}{22.4}\times 10=0.446$ moles

From the equation:

$2$ moles of $H_{2}S$ gas reacts with $1$ mole of $S{O}_2$ gas

$0.446$ moles of $H_{2}S$ gas reacts with $\frac{1}{2}\times 0.446=0.226$ moles of $S{O}_2$ gas

But we have excess of $S{O}_2$ gas that is $0.446$ moles of $S{O}_2$ gas is present.

So, $H_{2}S$ is the limiting reagent.

Now $0.446$ moles of $H_{2}S$ gas will consume $0.223$ moles of $S{O}_2$ gas and all $H_{2}S$ gas will be consumed.

Remaining moles of $H_{2}S$ gas after the reaction = $0$ moles

Remaining moles of $S{O}_2$ gas after reaction = $0.446-0.223$ moles

= $0.223$ moles

Now at NTP, 1 mole of $S{O}_2$ gas = $22.4$ L

$\Rightarrow 0.223$ moles of $S{O}_2$ gas = $22.4\times 0.223$ L

= $5$ L

Thus after the completion of the reaction volume of gas will be $5$ L