Answer: The volume of hydrogen gas produced will be, 12.4 L
Explanation : Given,
Mass of
= 54.219 g
Number of atoms of
= 
Molar mass of
= 98 g/mol
First we have to calculate the moles of
and
.


and,

Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:

From the balanced reaction we conclude that
As, 3 mole of
react with 2 mole of 
So, 0.553 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of 
From the reaction, we conclude that
As, 3 mole of
react to give 3 mole of 
So, 0.553 mole of
react to give 0.553 mole of 
Now we have to calculate the volume of
gas at STP.
As we know that, 1 mole of substance occupies 22.4 L volume of gas.
As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas
So, 0.553 mole of hydrogen gas occupies
volume of hydrogen gas
Therefore, the volume of hydrogen gas produced will be, 12.4 L