Answer : The volume of [tex]O_2[/tex] gas needed are, 90 L
Explanation : Given,
Volume of [tex]H_2S[/tex] = 60 L
Now we have to determine the volume of [tex]O_2[/tex] needed.
As we know that at STP, 1 mole of gas contains 22.4 L volume of gas.
The given balanced chemical reaction is:
[tex]2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)[/tex]
By the stoichiometry we can say that, 2 moles of [tex]H_2S[/tex] react with 3 moles of oxygen gas to give 2 moles of [tex]SO_2[/tex] gas and 2 moles of water vapor.
From the balanced reaction we conclude that,
As, [tex]2\times 22.4L[/tex] volume of [tex]H_2S[/tex] react with [tex]3\times 22.4L[/tex] volume of [tex]O_2[/tex] gas
So, [tex]60L[/tex] volume of [tex]H_2S[/tex] react with [tex]\frac{3\times 22.4L}{2\times 22.4L}\times 60L=90L[/tex] volume of [tex]O_2[/tex] gas
Thus, the volume of [tex]O_2[/tex] gas needed are, 90 L
