What volume of hydrogen will be produced at STP by the reaction of 78.33g of aluminum with excess water

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