This problem can be solved in a simplest way, without using Ideal Gas Equation and lengthy calculations.
The balance chemical equation is,
2 Al + 6 H₂O → 2 Al(OH)₃ + 3 H₂
Remember at STP 1 mole of any gas exactly occupies 22.4 L volume.
So,
54 g (2 mole) Al on reaction produces = 67.2 L (3 mole) H₂
Then,
78.33 g of Al will produce = X L of H₂
Solving for X,
X = (78.33 g × 67.2 L) ÷ 54 g
X = 97.47 L of H₂ Gas
