The electrolysis of water forms H_(2) and O_(2) 2H_(2)Oarrow 2H_(2)+O_(2) What is the percent yield of O_(2) if 10.2 g of O_(2) is produced from the decomposition of 17.0 g of H_(2)O Use % Yield=(Actualyield)/(Theoreticalyield)times 100 15.1% 33.8% 60.1% 67.6%

To calculate the percent yield of $O_{2}$, we first need to determine the theoretical yield of $O_{2}$ based on the given amount of $H_{2}O$.<br /><br />Step 1: Calculate the molar mass of $H_{2}O$.<br />The molar mass of $H_{2}O$ is 18.01528 g/mol.<br /><br />Step 2: Calculate the number of moles of $H_{2}O$.<br />Number of moles of $H_{2}O$ = $\frac{17.0 \text{ g}}{18.01528 \text{ g/mol}}$ = 0.943 moles<br /><br />Step 3: Use the stoichiometry of the reaction to find the theoretical yield of $O_{2}$.<br />According to the balanced equation, 2 moles of $H_{2}O$ produce 1 mole of $O_{2}$. Therefore, 0.943 moles of $H_{2}O$ will produce 0.4715 moles of $O_{2}$.<br /><br />Step 4: Calculate the theoretical yield of $O_{2}$ in grams.<br />The molar mass of $O_{2}$ is 31.998 g/mol.<br />Theoretical yield of $O_{2}$ = 0.4715 moles * 31.998 g/mol = 15.1 g<br /><br />Step 5: Calculate the percent yield of $O_{2}$.<br />Percent yield = $\frac{10.2 \text{ g}}{15.1 \text{ g}} \times 100$ = 67.6%<br /><br />Therefore, the correct answer is $67.6\%$.