Question

The electrolysis of water forms H2 and O2. 2H2O mc017-1.jpg 2H2 + O2 What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O? Use mc017-2.jpg. 15.1% 33.8% 60.1% 67.6%

Answer 1

Answer : The percent yield of $O_2$ is, 67.819%

Solution : Given,

Mass of water = 17 g

Molar mass of water = 18 g/mole

Molar mass of $O_2$ = 32 g/mole

Experimental yield of $O_2$ = 10.2 g

First we have to calculate the moles of water.

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{17g}{18g/mole}=0.94moles$

Now we have to calculate the moles of $O_2$

The given balanced reaction is,

$2H_2O\rightarrow 2H_2+O_2$

From the reaction, we conclude that

2 moles of water decompose to give 1 mole of $O_2$

0.94 moles of water decompose to give $\frac{1}{2}\times 0.94=0.47$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(0.47moles)\times (32g/mole)=15.04g$

Theoretical yield of $O_2$ = 15.04 g

Now we have to calculate the percent yield of $O_2$

$\% \text{ yield of }O_2=\frac{\text{ Experimental yield of }O_2}{\text{ Theretical yield of }O_2}\times 100$

$\% \text{ yield of }O_2=\frac{10.2g}{15.04}\times 100=67.819\%$

Therefore, the percent yield of $O_2$ is, 67.819%