What is the limiting reagent and thus the mass of product for the reaction: P4O10 + 6H2O --> 4H3PO4 if 5.00 g of P4O10 react with 1.50 g of water?

1)   Determine the moles of each reactant and thus the limiting reagent:# moles= mass (g)/M_{r .} M_r of P₄O₁₀ = (4 x 30.97) + (10 x 16.00) = 283.88 g mol^-1. Moles of P₄O₁₀ = 5.00/283.88 = 1.76 x 10^-2 mol. M_r of H₂O = (2 x 1.01) + 16.00 = 18.02 g mol^-1. Moles of H₂O = 1.50/18.02 = 8.32 x 10^-2 mol. Since there are fewer moles of P₄O₁₀, it is the limiting reagent in the reaction. H₂O is in excess. 2)   Determine the theoretical mass of product according to the limiting reagent: 1 mole of P₄O₁₀ reacts to form 4 moles of H₃PO₄ product. Assuming that all the P₄O₁₀ reacts, then 4 x 1.76 x 10^-2 = 7.04 x 10^-2 mol of H₃PO₄ product are formed. Rearranging the equation in part 1: Mass (g) = # moles x M_r , M_r of H₃PO₄ = (3 x 1.01) + 30.97 + (4 x 16.00) = 98.00 g mol^-1 . Therefore, mass of H₃PO₄ = 7.04 x 10^-2 mol x 98.00 g mol^-1 = 6.90 g