Answer :
To determine the percentage yield of calcium nitride in this reaction, we need to follow these steps:
1. Find the moles of reactants:
- Calcium (Ca):
- The molar mass of calcium is 40.08 g/mol.
- Moles of calcium = 56.6 g ÷ 40.08 g/mol = 1.41 moles
- Nitrogen gas (N₂):
- The molar mass of nitrogen gas (N₂) is 28.02 g/mol.
- Moles of nitrogen = 30.5 g ÷ 28.02 g/mol = 1.09 moles
2. Determine the limiting reactant:
- From the balanced reaction [tex]\(3 \text{ Ca} + \text{ N}_2 \rightarrow \text{ Ca}_3\text{N}_2\)[/tex], we know that 3 moles of Ca react with 1 mole of N₂ to produce 1 mole of Ca₃N₂.
- Moles of Ca₃N₂ possible from Ca: [tex]\(1.41 \, \text{moles Ca} ÷ 3 = 0.47\)[/tex] moles
- Moles of Ca₃N₂ possible from N₂: [tex]\(1.09 \, \text{moles N}_2 ÷ 1 = 1.09\)[/tex] moles
- Therefore, calcium is the limiting reactant since it produces fewer moles of Ca₃N₂ (0.47 moles).
3. Calculate the theoretical yield:
- The theoretical moles of Ca₃N₂ that can be produced is 0.47 moles, based on the limiting reactant (Ca).
- The molar mass of calcium nitride ([tex]\(\text{Ca}_3\text{N}_2\)[/tex]) is 148.248 g/mol.
- Theoretical mass of Ca₃N₂ = 0.47 moles × 148.248 g/mol = 69.78 g
4. Calculate the percentage yield:
- Actual mass of Ca₃N₂ produced = 32.4 g
- Percentage yield = (Actual yield ÷ Theoretical yield) × 100%
- Percentage yield = (32.4 g ÷ 69.78 g) × 100% = 46.5%
Thus, the percentage yield of calcium nitride for this reaction is 46.5%. Therefore, the correct answer choice is c. 46.5%.
1. Find the moles of reactants:
- Calcium (Ca):
- The molar mass of calcium is 40.08 g/mol.
- Moles of calcium = 56.6 g ÷ 40.08 g/mol = 1.41 moles
- Nitrogen gas (N₂):
- The molar mass of nitrogen gas (N₂) is 28.02 g/mol.
- Moles of nitrogen = 30.5 g ÷ 28.02 g/mol = 1.09 moles
2. Determine the limiting reactant:
- From the balanced reaction [tex]\(3 \text{ Ca} + \text{ N}_2 \rightarrow \text{ Ca}_3\text{N}_2\)[/tex], we know that 3 moles of Ca react with 1 mole of N₂ to produce 1 mole of Ca₃N₂.
- Moles of Ca₃N₂ possible from Ca: [tex]\(1.41 \, \text{moles Ca} ÷ 3 = 0.47\)[/tex] moles
- Moles of Ca₃N₂ possible from N₂: [tex]\(1.09 \, \text{moles N}_2 ÷ 1 = 1.09\)[/tex] moles
- Therefore, calcium is the limiting reactant since it produces fewer moles of Ca₃N₂ (0.47 moles).
3. Calculate the theoretical yield:
- The theoretical moles of Ca₃N₂ that can be produced is 0.47 moles, based on the limiting reactant (Ca).
- The molar mass of calcium nitride ([tex]\(\text{Ca}_3\text{N}_2\)[/tex]) is 148.248 g/mol.
- Theoretical mass of Ca₃N₂ = 0.47 moles × 148.248 g/mol = 69.78 g
4. Calculate the percentage yield:
- Actual mass of Ca₃N₂ produced = 32.4 g
- Percentage yield = (Actual yield ÷ Theoretical yield) × 100%
- Percentage yield = (32.4 g ÷ 69.78 g) × 100% = 46.5%
Thus, the percentage yield of calcium nitride for this reaction is 46.5%. Therefore, the correct answer choice is c. 46.5%.