Answer: The volume of hydrogen gas produced will be, 12.4 L
Explanation : Given,
Mass of  = 54.219 g
 = 54.219 g
Number of atoms of  =
 = 
Molar mass of  = 98 g/mol
 = 98 g/mol
First we have to calculate the moles of  and
 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
 react with 2 mole of 
So, 0.553 moles of  react with
 react with  moles of
 moles of 
From this we conclude that,  is an excess reagent because the given moles are greater than the required moles and
 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.
 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
 react to give 3 mole of 
So, 0.553 mole of  react to give 0.553 mole of
 react to give 0.553 mole of 
Now we have to calculate the volume of  gas at STP.
  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
 volume of hydrogen gas
Therefore, the volume of hydrogen gas produced will be, 12.4 L