Question

The electrolysis of water forms H2 and O2. 2H2O 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

Answer 1

Answer: The percent yield of oxygen gas is 67.53 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of water = 17.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{17.0g}{18g/mol}=0.944mol$

For the given chemical equation:

$2H_2O\rightarrow 2H_2+O_2$

By Stoichiometry of the reaction:

2 moles of water produces 1 mole of oxygen gas

So, 0.944 moles of water will produce = $\frac{1}{2}\times 0.944=0.472moles$ of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.472 moles

Putting values in equation 1, we get:

$0.472mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.472mol\times 32g/mol)=15.104g$

To calculate the percentage yield of oxygen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of oxygen gas = 10.2 g

Theoretical yield of oxygen gas = 15.104 g

Putting values in above equation, we get:

$\%\text{ yield of oxygen gas}=\frac{10.2g}{15.104g}\times 100\\\\\% \text{yield of oxygen gas}=67.53\%$

Hence, the percent yield of oxygen gas is 67.53 %.

Answer 2

Answer:67.6

Explanation: On edg 2020