Limiting Reactant Problem

Question

What is the limiting reactant, what is the excess reactant and by how much and what is the maximum amount of Aluminum Phosphate produced if 25.0 g of Aluminum reacts with 100.0 grams of H3PO4 according to the reactions below. 

                                                    2 Al  +  2 H3PO4   3 H2  +  2 AlPO4

Answer 1

(25.0g Al) * (1 mole Al /26.98g Al) * (2 mole AlPO₄ /2 mole Al) = 0.926₆ mole AlPO₄.

(100.0g H₃PO₄) * (1 mole H₃PO₄ /97.99g H₃PO₄) * (2 mole AlPO₄ /2 mole H₃PO₄) = 1.020₅ mole AlPO₄.

Limiting reactant is Al, and excess is H₃PO₄. Calculate excess amount from excess moles above:

(Δmole = 0.094 mole AlPO₄) * (2 mole H₃PO₄ /2 mole AlPO₄) * (97.99g H₃PO₄ /1 mole H₃PO₄) = 9.2g H₃PO₄.