This is just what we covered in chemistry class-so let me see how to solve thisFor any stoichiometry problem it's imperative we have a balanced equation to work withWhat do we know? We know Al reacts with HCl to yield AlCl3 (remember aluminum loses three electrons, always, to form a cation, so we must have 3 atoms of chlorine for one atom of aluminumAnd they tell us that H2 is producedSo so far we have: Al + HCl → AlCl3 + H2 Now to balancing this equationNever start balancing with hydrogen or oxygenSo we look at Al, and there is 1 mol of aluminum in both the reactants (R) and products (P)So it's balancedBut Cl isn't, right? What if we change the coefficient of AlCl3 to two, and have 2 mols of Cl3 in the products (236 Cl total)Then, change the coefficient of HCl in the R to 6 as well, and balance Al in the R to reflect that (2 Al) and, finally, H2 to reflect that (326) 2Al + 6HCl → 2AlCl3 + 3H2 Now you're ready to begin your calculations! Let's convert grams of HCl to mols, so we can use the mole ratios from our balanced chemical equation (BCE)We take our molar masses from the periodic table to two decimal placesSo 1 H(1.01 g/mol H) + 1 Cl(35.45 g/mol Cl) equals 36.46 g/mol HClNow we can begin dimensional analysisBegin with the information we're given-73 g AlMultiply it by 1 mol HCl over 36.46 g HCl (the molar mass we calculated), so we can cross out grams HClThen relate it to 2 mol Al over 6 mol HCl (from our BCE) so mols HCl can cancelNow we're left with mols Al and one more step is needed-get to grams AlRelate 1 mol of Al to its mass we get straight from the periodic table-26.98 g AlCross out mols Al and it's all down to simple arithmetic! I got 18 grams Al, rounded to 2 sig figs.
