Answer :
To determine the percent yield of the reaction, we can follow these steps:
1. Balance the Chemical Reaction:
The balanced chemical equation for the reaction is:
[tex]\[ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \][/tex]
2. Calculate Moles of Reactants:
- Hydrogen ([tex]\(\text{H}_2\)[/tex]):
The molar mass of [tex]\(\text{H}_2\)[/tex] is [tex]\(2 \, \text{g/mol}\)[/tex].
Moles of [tex]\(\text{H}_2\)[/tex] = [tex]\(\frac{10.0 \, \text{g}}{2 \, \text{g/mol}} = 5.0 \, \text{moles}\)[/tex]
- Oxygen ([tex]\(\text{O}_2\)[/tex]):
The molar mass of [tex]\(\text{O}_2\)[/tex] is [tex]\(32 \, \text{g/mol}\)[/tex].
Moles of [tex]\(\text{O}_2\)[/tex] = [tex]\(\frac{10.0 \, \text{g}}{32 \, \text{g/mol}} = 0.3125 \, \text{moles}\)[/tex]
3. Determine the Limiting Reactant:
From the balanced equation, 2 moles of [tex]\(\text{H}_2\)[/tex] react with 1 mole of [tex]\(\text{O}_2\)[/tex].
- If we have 5.0 moles of [tex]\(\text{H}_2\)[/tex], it would need 2.5 moles of [tex]\(\text{O}_2\)[/tex] to react completely.
- We only have 0.3125 moles of [tex]\(\text{O}_2\)[/tex], thus [tex]\(\text{O}_2\)[/tex] is the limiting reactant.
4. Calculate Theoretical Yield of [tex]\(\text{H}_2\text{O}\)[/tex]:
- Since [tex]\(\text{O}_2\)[/tex] is the limiting reactant, it limits the amount of [tex]\(\text{H}_2\text{O}\)[/tex] that can be produced.
- According to the equation, 1 mole of [tex]\(\text{O}_2\)[/tex] produces 2 moles of [tex]\(\text{H}_2\text{O}\)[/tex]. Thus, 0.3125 moles of [tex]\(\text{O}_2\)[/tex] produce:
[tex]\(0.3125 \times 2 = 0.625 \, \text{moles of } \text{H}_2\text{O}\)[/tex]
- The molar mass of [tex]\(\text{H}_2\text{O}\)[/tex] is [tex]\(18 \, \text{g/mol}\)[/tex].
Theoretical mass of [tex]\(\text{H}_2\text{O}\)[/tex] = [tex]\(0.625 \times 18 = 11.25 \, \text{g}\)[/tex]
5. Calculate Percent Yield:
- The actual yield of [tex]\(\text{H}_2\text{O}\)[/tex] is [tex]\(7.81 \, \text{g}\)[/tex].
- Percent yield = [tex]\(\left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100\)[/tex]
Percent yield = [tex]\(\frac{7.81 \, \text{g}}{11.25 \, \text{g}} \times 100 = 69.4\%\)[/tex]
Thus, the percent yield of the reaction is [tex]\(69.4\%\)[/tex]. The correct answer is [tex]\(c\)[/tex] [tex]\(69.4\%\)[/tex].
1. Balance the Chemical Reaction:
The balanced chemical equation for the reaction is:
[tex]\[ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \][/tex]
2. Calculate Moles of Reactants:
- Hydrogen ([tex]\(\text{H}_2\)[/tex]):
The molar mass of [tex]\(\text{H}_2\)[/tex] is [tex]\(2 \, \text{g/mol}\)[/tex].
Moles of [tex]\(\text{H}_2\)[/tex] = [tex]\(\frac{10.0 \, \text{g}}{2 \, \text{g/mol}} = 5.0 \, \text{moles}\)[/tex]
- Oxygen ([tex]\(\text{O}_2\)[/tex]):
The molar mass of [tex]\(\text{O}_2\)[/tex] is [tex]\(32 \, \text{g/mol}\)[/tex].
Moles of [tex]\(\text{O}_2\)[/tex] = [tex]\(\frac{10.0 \, \text{g}}{32 \, \text{g/mol}} = 0.3125 \, \text{moles}\)[/tex]
3. Determine the Limiting Reactant:
From the balanced equation, 2 moles of [tex]\(\text{H}_2\)[/tex] react with 1 mole of [tex]\(\text{O}_2\)[/tex].
- If we have 5.0 moles of [tex]\(\text{H}_2\)[/tex], it would need 2.5 moles of [tex]\(\text{O}_2\)[/tex] to react completely.
- We only have 0.3125 moles of [tex]\(\text{O}_2\)[/tex], thus [tex]\(\text{O}_2\)[/tex] is the limiting reactant.
4. Calculate Theoretical Yield of [tex]\(\text{H}_2\text{O}\)[/tex]:
- Since [tex]\(\text{O}_2\)[/tex] is the limiting reactant, it limits the amount of [tex]\(\text{H}_2\text{O}\)[/tex] that can be produced.
- According to the equation, 1 mole of [tex]\(\text{O}_2\)[/tex] produces 2 moles of [tex]\(\text{H}_2\text{O}\)[/tex]. Thus, 0.3125 moles of [tex]\(\text{O}_2\)[/tex] produce:
[tex]\(0.3125 \times 2 = 0.625 \, \text{moles of } \text{H}_2\text{O}\)[/tex]
- The molar mass of [tex]\(\text{H}_2\text{O}\)[/tex] is [tex]\(18 \, \text{g/mol}\)[/tex].
Theoretical mass of [tex]\(\text{H}_2\text{O}\)[/tex] = [tex]\(0.625 \times 18 = 11.25 \, \text{g}\)[/tex]
5. Calculate Percent Yield:
- The actual yield of [tex]\(\text{H}_2\text{O}\)[/tex] is [tex]\(7.81 \, \text{g}\)[/tex].
- Percent yield = [tex]\(\left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100\)[/tex]
Percent yield = [tex]\(\frac{7.81 \, \text{g}}{11.25 \, \text{g}} \times 100 = 69.4\%\)[/tex]
Thus, the percent yield of the reaction is [tex]\(69.4\%\)[/tex]. The correct answer is [tex]\(c\)[/tex] [tex]\(69.4\%\)[/tex].
