Question

46.

How many liters of O2 are needed to react completely with 60 L of H2S at STP (please show your work)


2H2S(g) + 3O2(g)—>2SO2(g) + 2H2O(g)

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* 22.4L` volume of
`H_2S` react with
`3* 22.4L` volume of
`O_2` gas

So,
`60L` volume of
`H_2S` react with
`(3* 22.4L)/(2* 22.4L)* 60L=90L` volume of
`O_2` gas

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