Question

46.

Answer 1

Answer : The volume of $O_2$ gas needed are, 90 L

Explanation : Given,

Volume of $H_2S$ = 60 L

Now we have to determine the volume of $O_2$ needed.

As we know that at STP, 1 mole of gas contains 22.4 L volume of gas.

The given balanced chemical reaction is:

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$

By the stoichiometry we can say that, 2 moles of $H_2S$ react with 3 moles of oxygen gas to give 2 moles of $SO_2$ gas and 2 moles of water vapor.

From the balanced reaction we conclude that,

As, $2\times 22.4L$ volume of $H_2S$ react with $3\times 22.4L$ volume of $O_2$ gas

So, $60L$ volume of $H_2S$ react with $\frac{3\times 22.4L}{2\times 22.4L}\times 60L=90L$ volume of $O_2$ gas

Thus, the volume of $O_2$ gas needed are, 90 L