To solve the problem, we need to go through several steps, including calculating the number of moles of hydrogen gas (H₂), the theoretical yield of water (H₂O), and finally the percent yield of the reaction.
1. Calculate the number of moles of H₂:
Given:
- Mass of H₂ = 5.58 g
- Molar mass of H₂ = 2.016 g/mol
[tex]\[
\text{Moles of H₂} = \frac{\text{Mass of H₂}}{\text{Molar mass of H₂}} = \frac{5.58 \text{ g}}{2.016 \text{ g/mol}} = 2.767857142857143 \text{ moles}
\][/tex]
2. Determine the moles of H₂O produced:
According to the balanced chemical equation:
[tex]\[
2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l)
\][/tex]
The stoichiometric ratio between H₂ and H₂O is 1:1. Therefore, the moles of H₂O produced will be the same as the moles of H₂:
[tex]\[
\text{Moles of H₂O} = 2.767857142857143 \text{ moles}
\][/tex]
3. Calculate the theoretical yield of H₂O:
Given:
- Molar mass of H₂O = 18.015 g/mol
[tex]\[
\text{Theoretical yield of H₂O} = \text{Moles of H₂O} \times \text{Molar mass of H₂O} = 2.767857142857143 \text{ moles} \times 18.015 \text{ g/mol} = 49.862946428571426 \text{ g}
\][/tex]
4. Calculate the percent yield:
Given:
- Actual yield of H₂O = 32.8 g
- Theoretical yield of H₂O = 49.862946428571426 g
[tex]\[
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{32.8 \text{ g}}{49.862946428571426 \text{ g}} \right) \times 100 = 65.78030852425846 \%
\][/tex]
So, the correct answer is:
[tex]\[
65.7 \%
\][/tex]
This corresponds to the option:
[tex]\[
65.7 \%
\][/tex]
