1. The problem statement, all variables and given/known data what is the mass of alum that can be prepared from 1.340g of Al. 2. Relevant equations 2Al+2KOH+4H2SO4+22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O 3. The attempt at a solution given Al is the limiting reagent. Find the moles of Al using n=m/M I find Al moles to be 0.049662mol. Now multiply this by the molar ratio of 2/2 (which is 1) to find the moles of alum produced. now calculate for the mass, m = nM. The molar mass of alum is 474.3g/mol. I find the mass of alum to be 23.55g. Am I right?