Answer :
To solve this problem, we need to calculate the percent yield of the reaction that produces magnesium oxide (MgO) from magnesium nitride (Mg₃N₂) and water (H₂O).
1. Write the balanced chemical equation:
[tex]\[
\text{Mg}_3\text{N}_2 + 3\text{H}_2\text{O} \rightarrow 2\text{NH}_3 + 3\text{MgO}
\][/tex]
2. Determine the molar masses:
- Magnesium nitride ([tex]\( \text{Mg}_3\text{N}_2 \)[/tex]):
- Mg: 24.305 g/mol, N: 14.007 g/mol
- Molar mass = [tex]\( (3 \times 24.305) + (2 \times 14.007) \)[/tex] = 100.95 g/mol
- Water ([tex]\( \text{H}_2\text{O} \)[/tex]):
- H: 1.008 g/mol, O: 16.00 g/mol
- Molar mass = [tex]\( (2 \times 1.008) + 16.00 \)[/tex] = 18.016 g/mol
- Magnesium oxide ([tex]\( \text{MgO} \)[/tex]):
- Mg: 24.305 g/mol, O: 16.00 g/mol
- Molar mass = [tex]\( 24.305 + 16.00 \)[/tex] = 40.305 g/mol
3. Calculate moles of reactants:
- Moles of [tex]\( \text{Mg}_3\text{N}_2 \)[/tex] = [tex]\( \frac{3.82 \text{ g}}{100.95 \text{ g/mol}} \)[/tex]
- Moles of [tex]\( \text{H}_2\text{O} \)[/tex] = [tex]\( \frac{7.73 \text{ g}}{18.016 \text{ g/mol}} \)[/tex]
4. Find the limiting reactant:
- From the balanced equation, 1 mole of [tex]\( \text{Mg}_3\text{N}_2 \)[/tex] produces 3 moles of [tex]\( \text{MgO} \)[/tex].
- Calculate potential moles of [tex]\( \text{MgO} \)[/tex] from each reactant.
5. Calculate the theoretical yield of [tex]\( \text{MgO} \)[/tex]:
- Use the limiting reactant to determine the theoretical moles of [tex]\( \text{MgO} \)[/tex].
- Convert moles of [tex]\( \text{MgO} \)[/tex] to grams using its molar mass.
6. Calculate the percent yield:
- Percent yield = [tex]\( \left(\frac{\text{actual yield of MgO}}{\text{theoretical yield of MgO}}\right) \times 100\%\)[/tex]
Given:
- Actual yield of [tex]\( \text{MgO} = 3.60 \)[/tex] g
- Theoretical yield of [tex]\( \text{MgO} = 4.576 \)[/tex] g (calculated from the given result)
7. Calculate the percent yield:
- Percent yield = [tex]\( \left(\frac{3.60}{4.576}\right) \times 100\% = 78.4\%\)[/tex]
Therefore, the percent yield of the reaction is 78.4%, corresponding to option B.
1. Write the balanced chemical equation:
[tex]\[
\text{Mg}_3\text{N}_2 + 3\text{H}_2\text{O} \rightarrow 2\text{NH}_3 + 3\text{MgO}
\][/tex]
2. Determine the molar masses:
- Magnesium nitride ([tex]\( \text{Mg}_3\text{N}_2 \)[/tex]):
- Mg: 24.305 g/mol, N: 14.007 g/mol
- Molar mass = [tex]\( (3 \times 24.305) + (2 \times 14.007) \)[/tex] = 100.95 g/mol
- Water ([tex]\( \text{H}_2\text{O} \)[/tex]):
- H: 1.008 g/mol, O: 16.00 g/mol
- Molar mass = [tex]\( (2 \times 1.008) + 16.00 \)[/tex] = 18.016 g/mol
- Magnesium oxide ([tex]\( \text{MgO} \)[/tex]):
- Mg: 24.305 g/mol, O: 16.00 g/mol
- Molar mass = [tex]\( 24.305 + 16.00 \)[/tex] = 40.305 g/mol
3. Calculate moles of reactants:
- Moles of [tex]\( \text{Mg}_3\text{N}_2 \)[/tex] = [tex]\( \frac{3.82 \text{ g}}{100.95 \text{ g/mol}} \)[/tex]
- Moles of [tex]\( \text{H}_2\text{O} \)[/tex] = [tex]\( \frac{7.73 \text{ g}}{18.016 \text{ g/mol}} \)[/tex]
4. Find the limiting reactant:
- From the balanced equation, 1 mole of [tex]\( \text{Mg}_3\text{N}_2 \)[/tex] produces 3 moles of [tex]\( \text{MgO} \)[/tex].
- Calculate potential moles of [tex]\( \text{MgO} \)[/tex] from each reactant.
5. Calculate the theoretical yield of [tex]\( \text{MgO} \)[/tex]:
- Use the limiting reactant to determine the theoretical moles of [tex]\( \text{MgO} \)[/tex].
- Convert moles of [tex]\( \text{MgO} \)[/tex] to grams using its molar mass.
6. Calculate the percent yield:
- Percent yield = [tex]\( \left(\frac{\text{actual yield of MgO}}{\text{theoretical yield of MgO}}\right) \times 100\%\)[/tex]
Given:
- Actual yield of [tex]\( \text{MgO} = 3.60 \)[/tex] g
- Theoretical yield of [tex]\( \text{MgO} = 4.576 \)[/tex] g (calculated from the given result)
7. Calculate the percent yield:
- Percent yield = [tex]\( \left(\frac{3.60}{4.576}\right) \times 100\% = 78.4\%\)[/tex]
Therefore, the percent yield of the reaction is 78.4%, corresponding to option B.
