Question

The mass of $N_2F_4$ produced by the reaction of 2.0 g of $NH_3$ and 8.0 g of $F_2$ is 3.56 g. What is the percentage yield?

• A. 79.0
• B. 71.2
• C. 84.6
• D. None of these

Answer 1

Answer: (D)

None of these (calculated 81.3%)

Answer 2

1. Calculate Molar Masses:

   • $NH_3$: $14.01 + 3 \times 1.01 = 17.04$ g/mol
   • $F_2$: $2 \times 19.00 = 38.00$ g/mol
   • $N_2F_4$: $2 \times 14.01 + 4 \times 19.00 = 28.02 + 76.00 = 104.02$ g/mol

2. Calculate Moles of Reactants:

   • Moles of $NH_3 = \frac{2.0 \text{ g}}{17.04 \text{ g/mol}} \approx 0.1174$ mol
   • Moles of $F_2 = \frac{8.0 \text{ g}}{38.00 \text{ g/mol}} \approx 0.2105$ mol

3. Determine the Limiting Reactant: The balanced equation is $2NH_3 + 5F_2 \longrightarrow N_2F_4 + 6HF$. The stoichiometric ratio of $NH_3$ to $F_2$ is $2:5$.

   • To react completely with 0.1174 mol of $NH_3$, we need $0.1174 \times \frac{5}{2} = 0.2935$ mol of $F_2$. We only have 0.2105 mol of $F_2$.
   • To react completely with 0.2105 mol of $F_2$, we need $0.2105 \times \frac{2}{5} = 0.0842$ mol of $NH_3$. We have 0.1174 mol of $NH_3$. Since we have less $F_2$ than required to react with all $NH_3$, $F_2$ is the limiting reactant.

4. Calculate Theoretical Yield of $N_2F_4$: From the stoichiometry, 5 moles of $F_2$ produce 1 mole of $N_2F_4$. Moles of $N_2F_4$ produced = $0.2105 \text{ mol } F_2 \times \frac{1 \text{ mol } N_2F_4}{5 \text{ mol } F_2} = 0.0421$ mol $N_2F_4$. Theoretical yield of $N_2F_4 = 0.0421 \text{ mol} \times 104.02 \text{ g/mol} \approx 4.379$ g.

5. Calculate Percentage Yield: Percentage Yield $= \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100$ Percentage Yield $= \left( \frac{3.56 \text{ g}}{4.379 \text{ g}} \right) \times 100 \approx 81.3\%$

   Note: The calculated value of 81.3% is not directly among the options. Therefore, the correct option is (D) None of these.