Question

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

collected after the experiment?
H20 --> H2 + O2

Answer 1

Approximately
`64\%`.

Step-by-step explanation:

`\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} * 100\%`

The actual yield of
`\rm O_2` was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

`\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)`.

Note the ratio between the coefficient of
`\rm H_2O\, (g)` and
`\rm O_2\, (g)`:

`\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = (1)/(2)`.

This ratio will be useful for finding the theoretical yield of
`\rm O_2\, (g)`.

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

• 
  `\rm H`:
  `1.008`.
• 
  `\rm O`:
  `15.999`.

Calculate the formula mass of
`\rm H_2O` and
`\rm O_2`:

`M(\mathrm{H_2O}) =2* 1.008 + 15.999 = 18.015\; \rm g \cdot mol^(-1)`.

`M(\mathrm{O_2}) =2* 15.999 = 31.998\; \rm g \cdot mol^(-1)`.

Calculate the number of moles of molecules in
`9.8\; \rm g` of
`\rm H_2O`:

`\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = (9.8\; \rm g)/(18.015\; \rm g \cdot mol^(-1)) \approx 0.543991\;\rm g \cdot mol^(-1)`.

Make use of the ratio
`\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = (1)/(2)` to find the theoretical yield of
`\rm O_2` (in terms of number of moles of molecules.)

`\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx (1)/(2) * 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}`.

Calculate the mass of that approximately
`0.271996\; \rm mol` of
`\rm O_2` (theoretical yield.)

`\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol * 31.998\; \rm g \cdot mol^(-1) \approx 8.70331 \; \rm g \end{aligned}`.

That would correspond to the theoretical yield of
`\rm O_2` (in term of the mass of the product.)

Given that the actual yield is
`5.6\; \rm g`, calculate the percentage yield:

`\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} * 100\% \\ &\approx (5.6\; \rm g)/(8.70331\; \rm g) * 100\% \approx 64\%\end{aligned}`.