Respuesta :
Answer:
Approximately [tex]64\%[/tex].
Explanation:
[tex]\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%[/tex]
The actual yield of [tex]\rm O_2[/tex] was given. The theoretical yield needs to be calculated from the quantity of the reactant.
Balance the equation for the hydrolysis of water:
[tex]\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)[/tex].
Note the ratio between the coefficient of [tex]\rm H_2O\, (g)[/tex] and [tex]\rm O_2\, (g)[/tex]:
[tex]\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}[/tex].
This ratio will be useful for finding the theoretical yield of [tex]\rm O_2\, (g)[/tex].
Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.
- [tex]\rm H[/tex]: [tex]1.008[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
Calculate the formula mass of [tex]\rm H_2O[/tex] and [tex]\rm O_2[/tex]:
[tex]M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}[/tex].
Calculate the number of moles of molecules in [tex]9.8\; \rm g[/tex] of [tex]\rm H_2O[/tex]:
[tex]\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}[/tex].
Make use of the ratio [tex]\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}[/tex] to find the theoretical yield of [tex]\rm O_2[/tex] (in terms of number of moles of molecules.)
[tex]\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}[/tex].
Calculate the mass of that approximately [tex]0.271996\; \rm mol[/tex] of [tex]\rm O_2[/tex] (theoretical yield.)
[tex]\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}[/tex].
That would correspond to the theoretical yield of [tex]\rm O_2[/tex] (in term of the mass of the product.)
Given that the actual yield is [tex]5.6\; \rm g[/tex], calculate the percentage yield:
[tex]\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}[/tex].
