11. For the following reaction: P_(2)O_(5)(s)+3H_(2)O(l)arrow 2H_(3)PO_(4)(aq) 288 g of P_(2)O_(5) and 54 g of H_(2)O were supplied. a) Which is the limiting reactant? b) What is the theoretical yield of H_(3)PO_(4)

a) To determine the limiting reactant, we need to calculate the number of moles of each reactant and compare them to the stoichiometry of the reaction.First, let's calculate the number of moles of :Molar mass of = 2(30.97) + 5(16.00) = 141.94 g/molNumber of moles of = 288 g / 141.94 g/mol = 2.03 molNext, let's calculate the number of moles of :Molar mass of = 2(1.01) + 16.00 = 18.02 g/molNumber of moles of = 54 g / 18.02 g/mol = 3.00 molNow, let's compare the number of moles of each reactant to the stoichiometry of the reaction. According to the balanced equation, 1 mole of reacts with 3 moles of to produce 2 moles of .Since we have 2.03 moles of and 3.00 moles of , we can see that is the limiting reactant because it has fewer moles available for the reaction.b) To calculate the theoretical yield of , we need to use the stoichiometry of the reaction and the number of moles of the limiting reactant ( ).According to the balanced equation, 1 mole of produces 2 moles of . Therefore, 2.03 moles of will produce 2.03 x 2 = 4.06 moles of .Now, let's calculate the molar mass of :Molar mass of = 3(1.01) + 30.97 + 4(16.00) = 97.99 g/molFinally, let's calculate the theoretical yield in grams:Theoretical yield = 4.06 moles x 97.99 g/mol = 398.4 gTherefore, the theoretical yield of is 398.4 g.