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)/Mr . Mr 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. Mr 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 Mr , Mr 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
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