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Answer :
To determine the percent yield of MgO in the reaction, we need to follow these steps:
1. Calculate Moles of Reactants:
- Molar masses:
- Molar mass of [tex]\(Mg_3N_2\)[/tex] is approximately 100.95 g/mol.
- Molar mass of [tex]\(H_2O\)[/tex] is approximately 18.015 g/mol.
- Moles of [tex]\(Mg_3N_2\)[/tex]:
[tex]\[
\text{Moles of }Mg_3N_2 = \frac{\text{Mass of } Mg_3N_2}{\text{Molar mass of } Mg_3N_2} = \frac{3.82 \text{ g}}{100.95 \text{ g/mol}} \approx 0.0378 \text{ mol}
\][/tex]
- Moles of [tex]\(H_2O\)[/tex]:
[tex]\[
\text{Moles of }H_2O = \frac{\text{Mass of } H_2O}{\text{Molar mass of } H_2O} = \frac{7.73 \text{ g}}{18.015 \text{ g/mol}} \approx 0.4291 \text{ mol}
\][/tex]
2. Determine Limiting Reactant:
- According to the balanced equation [tex]\(Mg_3N_2 + 3H_2O \rightarrow 2NH_3 + 3MgO\)[/tex], 1 mole of [tex]\(Mg_3N_2\)[/tex] reacts with 3 moles of [tex]\(H_2O\)[/tex].
- Necessary moles of [tex]\(Mg_3N_2\)[/tex] to react with available [tex]\(H_2O\)[/tex]:
[tex]\[
\text{Moles of } Mg_3N_2 = \frac{0.4291 \text{ mol } H_2O}{3} \approx 0.143 \text{ mol}
\][/tex]
- Compare with available moles of [tex]\(Mg_3N_2\)[/tex] (0.0378 mol):
Since 0.0378 mol < 0.143 mol, [tex]\(Mg_3N_2\)[/tex] is the limiting reactant.
3. Calculate Theoretical Yield of [tex]\(MgO\)[/tex]:
- From the balanced equation, 1 mole of [tex]\(Mg_3N_2\)[/tex] produces 3 moles of [tex]\(MgO\)[/tex].
- Moles of [tex]\(MgO\)[/tex] produced theoretically:
[tex]\[
\text{Moles of } MgO = 3 \times \text{Moles of } Mg_3N_2 = 3 \times 0.0378 \approx 0.1135 \text{ mol}
\][/tex]
4. Convert Theoretical Moles of [tex]\(MgO\)[/tex] to Grams:
- Molar mass of [tex]\(MgO\)[/tex] is approximately 40.304 g/mol.
- Theoretical mass of [tex]\(MgO\)[/tex]:
[tex]\[
\text{Mass of } MgO = 0.1135 \text{ mol} \times 40.304 \text{ g/mol} \approx 4.576 \text{ g}
\][/tex]
5. Calculate Percent Yield:
- Actual yield of [tex]\(MgO\)[/tex] is given as 3.60 g.
- Percent yield is calculated as:
[tex]\[
\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100 = \left(\frac{3.60}{4.576}\right) \times 100 \approx 78.7\%
\][/tex]
Thus, the percent yield of the reaction is approximately 78.7%, so the correct answer is B) 78.7.
1. Calculate Moles of Reactants:
- Molar masses:
- Molar mass of [tex]\(Mg_3N_2\)[/tex] is approximately 100.95 g/mol.
- Molar mass of [tex]\(H_2O\)[/tex] is approximately 18.015 g/mol.
- Moles of [tex]\(Mg_3N_2\)[/tex]:
[tex]\[
\text{Moles of }Mg_3N_2 = \frac{\text{Mass of } Mg_3N_2}{\text{Molar mass of } Mg_3N_2} = \frac{3.82 \text{ g}}{100.95 \text{ g/mol}} \approx 0.0378 \text{ mol}
\][/tex]
- Moles of [tex]\(H_2O\)[/tex]:
[tex]\[
\text{Moles of }H_2O = \frac{\text{Mass of } H_2O}{\text{Molar mass of } H_2O} = \frac{7.73 \text{ g}}{18.015 \text{ g/mol}} \approx 0.4291 \text{ mol}
\][/tex]
2. Determine Limiting Reactant:
- According to the balanced equation [tex]\(Mg_3N_2 + 3H_2O \rightarrow 2NH_3 + 3MgO\)[/tex], 1 mole of [tex]\(Mg_3N_2\)[/tex] reacts with 3 moles of [tex]\(H_2O\)[/tex].
- Necessary moles of [tex]\(Mg_3N_2\)[/tex] to react with available [tex]\(H_2O\)[/tex]:
[tex]\[
\text{Moles of } Mg_3N_2 = \frac{0.4291 \text{ mol } H_2O}{3} \approx 0.143 \text{ mol}
\][/tex]
- Compare with available moles of [tex]\(Mg_3N_2\)[/tex] (0.0378 mol):
Since 0.0378 mol < 0.143 mol, [tex]\(Mg_3N_2\)[/tex] is the limiting reactant.
3. Calculate Theoretical Yield of [tex]\(MgO\)[/tex]:
- From the balanced equation, 1 mole of [tex]\(Mg_3N_2\)[/tex] produces 3 moles of [tex]\(MgO\)[/tex].
- Moles of [tex]\(MgO\)[/tex] produced theoretically:
[tex]\[
\text{Moles of } MgO = 3 \times \text{Moles of } Mg_3N_2 = 3 \times 0.0378 \approx 0.1135 \text{ mol}
\][/tex]
4. Convert Theoretical Moles of [tex]\(MgO\)[/tex] to Grams:
- Molar mass of [tex]\(MgO\)[/tex] is approximately 40.304 g/mol.
- Theoretical mass of [tex]\(MgO\)[/tex]:
[tex]\[
\text{Mass of } MgO = 0.1135 \text{ mol} \times 40.304 \text{ g/mol} \approx 4.576 \text{ g}
\][/tex]
5. Calculate Percent Yield:
- Actual yield of [tex]\(MgO\)[/tex] is given as 3.60 g.
- Percent yield is calculated as:
[tex]\[
\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100 = \left(\frac{3.60}{4.576}\right) \times 100 \approx 78.7\%
\][/tex]
Thus, the percent yield of the reaction is approximately 78.7%, so the correct answer is B) 78.7.
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