I balanced the equation as:
2Al(s)+6HCl(aq)-->2AlCl3(aq)+3H2(g)------assuming I this correctly :)
What mass of H2(g) is produced from the reaction of .75 g Al(s) with excess hydrochloric acid?
convert Al in to moles
now from equation you know
2 moles of Al gives you 3 moles of H2
start with moles of Al X 3moles of H2 / 2 moles of Al X molar mass of H2 / 1 mole of H2