We completed a lab: % yield of Alum (molar mass = 474.44 g/mol). If the limiting reactant continued to be Al powder, but we started with less than we were supposed to (0.897 g Al powder). We performed the lab anyway and actually still made 14.45 grams of alum. What was our percent yield? 2Al (s) + KOH(aq) +4H2SO4(aq) + 10H2O(l) → 2KAl(SO4)212H2O(cr) + 3H2(g)