**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**