11. For the following reaction: P_(2)O_(5)(s)+3H_(2)O(t)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?

To determine the limiting reactant, we need to compare the amount of each reactant available to the amount required by the balanced chemical equation.<br /><br />The balanced chemical equation is:<br />$P_{2}O_{5}(s)+3H_{2}O(t)\rightarrow 2H_{3}PO_{4}(aq)$<br /><br />Given:<br />- 288 g of $P_{2}O_{5}$<br />- 54 g of $H_{2}O$<br /><br />First, we need to calculate the molar masses of $P_{2}O_{5}$ and $H_{2}O$:<br />- Molar mass of $P_{2}O_{5}$ = 2(30.97) + 5(16.00) = 141.94 g/mol<br />- Molar mass of $H_{2}O$ = 2(1.01) + 16.00 = 18.02 g/mol<br /><br />Next, we convert the masses of $P_{2}O_{5}$ and $H_{2}O$ to moles:<br />- Moles of $P_{2}O_{5}$ = $\frac{288 \text{ g}}{141.94 \text{ g/mol}}$ = 2.03 moles<br />- Moles of $H_{2}O$ = $\frac{54 \text{ g}}{18.02 \text{ g/mol}}$ = 3.00 moles<br /><br />Now, we compare the mole ratio of the reactants to the stoichiometric ratio in the balanced equation:<br />- The balanced equation requires 1 mole of $P_{2}O_{5}$ for every 3 moles of $H_{2}O$.<br /><br />We have:<br />- 2.03 moles of $P_{2}O_{5}$<br />- 3.00 moles of $H_{2}O$<br /><br />Since we have more than enough $H_{2}O$ (3.00 moles) to react with 2.03 moles of $P_{2}O_{5}$, $P_{2}O_{5}$ is the limiting reactant.<br /><br />Therefore, the limiting reactant is $P_{2}O_{5}$.