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.<br /><br />First, let's calculate the number of moles of $P_{2}O_{5}$:<br />Molar mass of $P_{2}O_{5}$ = 2(30.97) + 5(16.00) = 141.94 g/mol<br />Number of moles of $P_{2}O_{5}$ = 288 g / 141.94 g/mol = 2.03 mol<br /><br />Next, let's calculate the number of moles of $H_{2}O$:<br />Molar mass of $H_{2}O$ = 2(1.01) + 16.00 = 18.02 g/mol<br />Number of moles of $H_{2}O$ = 54 g / 18.02 g/mol = 3.00 mol<br /><br />Now, let's compare the number of moles of each reactant to the stoichiometry of the reaction. According to the balanced equation, 1 mole of $P_{2}O_{5}$ reacts with 3 moles of $H_{2}O$ to produce 2 moles of $H_{3}PO_{4}$.<br /><br />Since we have 2.03 moles of $P_{2}O_{5}$ and 3.00 moles of $H_{2}O$, we can see that $P_{2}O_{5}$ is the limiting reactant because it has fewer moles available for the reaction.<br /><br />b) To calculate the theoretical yield of $H_{3}PO_{4}$, we need to use the stoichiometry of the reaction and the number of moles of the limiting reactant ($P_{2}O_{5}$).<br /><br />According to the balanced equation, 1 mole of $P_{2}O_{5}$ produces 2 moles of $H_{3}PO_{4}$. Therefore, 2.03 moles of $P_{2}O_{5}$ will produce 2.03 x 2 = 4.06 moles of $H_{3}PO_{4}$.<br /><br />Now, let's calculate the molar mass of $H_{3}PO_{4}$:<br />Molar mass of $H_{3}PO_{4}$ = 3(1.01) + 30.97 + 4(16.00) = 97.99 g/mol<br /><br />Finally, let's calculate the theoretical yield in grams:<br />Theoretical yield = 4.06 moles x 97.99 g/mol = 398.4 g<br /><br />Therefore, the theoretical yield of $H_{3}PO_{4}$ is 398.4 g.