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Sagot :
To determine the percent yield of lithium hydroxide (LiOH) from the reaction of lithium (Li) with water (H[tex]\(_2\)[/tex]O), follow these steps:
1. Identify the given data:
- Mass of lithium: [tex]\( 7.40 \text{ g} \)[/tex]
- Mass of water: [tex]\( 10.2 \text{ g} \)[/tex]
- Actual yield of lithium hydroxide: [tex]\( 12.1 \text{ g} \)[/tex]
- Molar masses (needed for calculations):
- Lithium ([tex]\( \text{Li} \)[/tex]): [tex]\( 6.94 \text{ g/mol} \)[/tex]
- Water ([tex]\( \text{H}_2\text{O} \)[/tex]): [tex]\( 18.0 \text{ g/mol} \)[/tex]
- Lithium hydroxide ([tex]\( \text{LiOH} \)[/tex]): [tex]\( 23.94 \text{ g/mol} \)[/tex]
2. Calculate the moles of reactants (Li and H[tex]\(_2\)[/tex]O):
[tex]\[ \text{Moles of Li} = \frac{\text{Mass of Li}}{\text{Molar mass of Li}} = \frac{7.40 \text{ g}}{6.94 \text{ g/mol}} \approx 1.066 \text{ moles} \][/tex]
[tex]\[ \text{Moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} = \frac{10.2 \text{ g}}{18.0 \text{ g/mol}} \approx 0.567 \text{ moles} \][/tex]
3. Determine the theoretical moles of lithium hydroxide (LiOH) produced:
According to the balanced chemical equation:
[tex]\[ 2 \text{ Li} + 2 \text{ H}_2\text{O} \rightarrow 2 \text{ LiOH} + \text{ H}_2 \][/tex]
The reaction stoichiometry shows that 2 moles of Li react with 2 moles of H[tex]\(_2\)[/tex]O to produce 2 moles of LiOH. Therefore, moles of LiOH produced should be equivalent to the moles of Li reacted considering lithium is the limiting reactant.
[tex]\[ \text{Theoretical moles of LiOH} = \text{Moles of Li} \approx 1.066 \text{ moles} \][/tex]
4. Convert the theoretical moles of lithium hydroxide to grams:
[tex]\[ \text{Theoretical yield of LiOH} = \text{Theoretical moles of LiOH} \times \text{Molar mass of LiOH} = 1.066 \text{ moles} \times 23.94 \text{ g/mol} \approx 25.53 \text{ g} \][/tex]
5. Calculate the percent yield:
[tex]\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{12.1 \text{ g}}{25.53 \text{ g}} \right) \times 100 \approx 47.4\% \][/tex]
Thus, the percent yield of lithium hydroxide is [tex]\( 47.4\% \)[/tex].
The correct answer is [tex]\( 47.4\%\)[/tex].
1. Identify the given data:
- Mass of lithium: [tex]\( 7.40 \text{ g} \)[/tex]
- Mass of water: [tex]\( 10.2 \text{ g} \)[/tex]
- Actual yield of lithium hydroxide: [tex]\( 12.1 \text{ g} \)[/tex]
- Molar masses (needed for calculations):
- Lithium ([tex]\( \text{Li} \)[/tex]): [tex]\( 6.94 \text{ g/mol} \)[/tex]
- Water ([tex]\( \text{H}_2\text{O} \)[/tex]): [tex]\( 18.0 \text{ g/mol} \)[/tex]
- Lithium hydroxide ([tex]\( \text{LiOH} \)[/tex]): [tex]\( 23.94 \text{ g/mol} \)[/tex]
2. Calculate the moles of reactants (Li and H[tex]\(_2\)[/tex]O):
[tex]\[ \text{Moles of Li} = \frac{\text{Mass of Li}}{\text{Molar mass of Li}} = \frac{7.40 \text{ g}}{6.94 \text{ g/mol}} \approx 1.066 \text{ moles} \][/tex]
[tex]\[ \text{Moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} = \frac{10.2 \text{ g}}{18.0 \text{ g/mol}} \approx 0.567 \text{ moles} \][/tex]
3. Determine the theoretical moles of lithium hydroxide (LiOH) produced:
According to the balanced chemical equation:
[tex]\[ 2 \text{ Li} + 2 \text{ H}_2\text{O} \rightarrow 2 \text{ LiOH} + \text{ H}_2 \][/tex]
The reaction stoichiometry shows that 2 moles of Li react with 2 moles of H[tex]\(_2\)[/tex]O to produce 2 moles of LiOH. Therefore, moles of LiOH produced should be equivalent to the moles of Li reacted considering lithium is the limiting reactant.
[tex]\[ \text{Theoretical moles of LiOH} = \text{Moles of Li} \approx 1.066 \text{ moles} \][/tex]
4. Convert the theoretical moles of lithium hydroxide to grams:
[tex]\[ \text{Theoretical yield of LiOH} = \text{Theoretical moles of LiOH} \times \text{Molar mass of LiOH} = 1.066 \text{ moles} \times 23.94 \text{ g/mol} \approx 25.53 \text{ g} \][/tex]
5. Calculate the percent yield:
[tex]\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{12.1 \text{ g}}{25.53 \text{ g}} \right) \times 100 \approx 47.4\% \][/tex]
Thus, the percent yield of lithium hydroxide is [tex]\( 47.4\% \)[/tex].
The correct answer is [tex]\( 47.4\%\)[/tex].
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