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

Sorry i dont understand the question can you rephase it?