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

The balanced equation for the above reaction is 
2Al + 6H₂O ---> 2Al(OH)₃ + 3H₂
stoichiometry of Al to H₂ is 2:3
number of Al moles reacted - 78.33 g / 27 g/mol = 2.901 mol
according to molar ratio 
2 mol of Al forms - 3 mol of H₂
therefore 2.901 mol of Al - forms 3/2 x 2.901 = 4.352 mol

molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP
if 1 mol occupies 22.4 L
then 4.352 mol occupies - 22.4 L/mol x 4.352 mol = 97.48 L
volume occupied by H₂ is 97.48 L