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
