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SCORPION-xisa [38]
3 years ago
13

Changing the Rate of Reaction

Chemistry
1 answer:
Lisa [10]3 years ago
3 0

give brainliest pls thanksss

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If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was
ANEK [815]

Answer:

Approximately 64\%.

Explanation:

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 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)})} = \frac{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\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}.

M(\mathrm{O_2}) =2\times 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})} = \frac{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)})} = \frac{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 \frac{1}{2} \times 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 \times 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}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}.

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