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Montano1993 [528]
3 years ago
13

46.

Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
3 0

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

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