the electrolysis of water forms h₂ and o₂.\n2h₂o → 2h₂ + o₂\nwhat is the percent yield of o₂ if 10.2 g of o₂…

the electrolysis of water forms h₂ and o₂.\n2h₂o → 2h₂ + o₂\nwhat is the percent yield of o₂ if 10.2 g of o₂ is produced from the decomposition of 17.0 g of h₂o? use %yield = \\frac{actual yield}{theoretical yield}×100.\n15.1%\n33.8%\n60.1%\n67.6%

the electrolysis of water forms h₂ and o₂.\n2h₂o → 2h₂ + o₂\nwhat is the percent yield of o₂ if 10.2 g of o₂ is produced from the decomposition of 17.0 g of h₂o? use %yield = \\frac{actual yield}{theoretical yield}×100.\n15.1%\n33.8%\n60.1%\n67.6%

Answer

Answer:

D. 67.6%

Explanation:

Step1: Calculate moles of water

$n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{17.0\ g}{18.02\ g/mol}\approx0.943\ mol$

Step2: Determine moles of oxygen from stoichiometry

From $2H_2O\rightarrow2H_2 + O_2$, mole - ratio of $H_2O$ to $O_2$ is 2:1. So $n_{O_2}=\frac{1}{2}n_{H_2O}=\frac{1}{2}\times0.943\ mol = 0.4715\ mol$

Step3: Calculate theoretical mass of oxygen

$m_{O_2 - theoretical}=n_{O_2}\times M_{O_2}=0.4715\ mol\times32.00\ g/mol = 15.09\ g$

Step4: Calculate percent - yield

$%Yield=\frac{Actual\ yield}{Theoretical\ yield}\times100=\frac{10.2\ g}{15.09\ g}\times100\approx67.6%$