text: Find the moles of Al:\n\nn = m/Ar = 10/27 = 0.37 mol (approx) where Ar = 27 is the relative atomic mass of Al.\n\nFind the moles of HCl:\n\nn = C*V = 6*0.1 = 0.6 mol HCl\n\nAccording to the balanced chem. equation:\n\n2Al + 6HCl  --> 2AlCl3 + 3H2\n\n1 mol of Al reacts with 3 mol of HCl, so\n0.37 mol of Al reacts with 3*0.37 = 1.11 mol of HCl\n\nSo HCl is the limiting reactant and because:\n\n6 mol of HCl produces 3 mol H2\n0.6 mol of HCl produces 0.3 mol H2\n\nThis quantity - 0.3 mol of H2 - at STP occupies:\n\nV = n*Vm = 0.3*22.4 = 6.72 L, where Vm = 22.4 L/mol is the molar volume in STP.