To determine the percent yield of [tex]\( \text{O}_2 \)[/tex] from the decomposition of [tex]\( \text{H}_2\text{O} \)[/tex], we will walk through the problem step-by-step, following the chemical reaction given:
[tex]\[ 2 \ \text{H}_2\text{O} \rightarrow 2 \ \text{H}_2 + \ \text{O}_2 \][/tex]
1. Given Data:
- Mass of [tex]\( \text{H}_2\text{O} \)[/tex]: [tex]\( 17.0 \ \text{grams} \)[/tex]
- Actual yield of [tex]\( \text{O}_2 \)[/tex]: [tex]\( 10.2 \ \text{grams} \)[/tex]
2. Find Molar Masses:
- Molar mass of [tex]\( \text{H}_2\text{O} \)[/tex]:
[tex]\[
\text{H}_2\text{O} = 2 \times 1.01 \ (\text{H}) + 16.00 \ (\text{O}) = 18.02 \ \text{g/mol}
\][/tex]
- Molar mass of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{O}_2 = 2 \times 16.00 \ (\text{O}) = 32.00 \ \text{g/mol}
\][/tex]
3. Calculate Moles of [tex]\( \text{H}_2\text{O} \)[/tex]:
[tex]\[
\text{Moles of } \text{H}_2\text{O} = \frac{\text{Mass of } \text{H}_2\text{O}}{\text{Molar mass of } \text{H}_2\text{O}} = \frac{17.0 \ \text{grams}}{18.02 \ \text{g/mol}} \approx 0.943 \ \text{moles}
\][/tex]
4. Determine Theoretical Moles of [tex]\( \text{O}_2 \)[/tex]:
According to the balanced equation, 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex] produce 1 mole of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{O}_2 = \frac{\text{Moles of } \text{H}_2\text{O}}{2} = \frac{0.943 \ \text{moles}}{2} \approx 0.472 \ \text{moles}
\][/tex]
5. Calculate Theoretical Yield of [tex]\( \text{O}_2 \)[/tex] in grams:
[tex]\[
\text{Theoretical yield of } \text{O}_2 = \text{Moles of } \text{O}_2 \times \text{Molar mass of } \text{O}_2 = 0.472 \ \text{moles} \times 32.00 \ \text{g/mol} \approx 15.09 \ \text{grams}
\][/tex]
6. Calculate Percent Yield:
[tex]\[
\text{Percent yield} = \left( \frac{\text{Actual yield of } \text{O}_2}{\text{Theoretical yield of } \text{O}_2} \right) \times 100 = \left( \frac{10.2 \ \text{grams}}{15.09 \ \text{grams}} \right) \times 100 \approx 67.6 \%
\][/tex]
Therefore, the percent yield of [tex]\( \text{O}_2 \)[/tex] is [tex]\( \boxed{67.6 \%} \)[/tex].
