To solve this problem, we need to follow a series of steps involving stoichiometry and the concept of molar mass. Here is a detailed, step-by-step solution:
1. Determine the molar mass of ammonia (NH₃):
- Nitrogen (N) has a molar mass of approximately 14 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.
- The formula for ammonia is NH₃, so its molar mass is:
[tex]\[
14 \, \text{g/mol (N)} + 3 \times 1 \, \text{g/mol (H)} = 17 \, \text{g/mol}
\][/tex]
2. Calculate the number of moles of ammonia (NH₃):
- Given mass of ammonia (NH₃) = 325 g
- Molar mass of NH₃ = 17 g/mol
- Number of moles is calculated using the formula:
[tex]\[
\text{moles of NH₃} = \frac{\text{mass of NH₃}}{\text{molar mass of NH₃}} = \frac{325 \, \text{g}}{17 \, \text{g/mol}} \approx 19.11 \, \text{moles}
\][/tex]
3. Use the stoichiometry of the reaction to determine moles of nitrogen gas (N₂) needed:
- According to the balanced chemical reaction:
[tex]\[
N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)
\][/tex]
- This means 1 mole of nitrogen gas (N₂) produces 2 moles of ammonia (NH₃).
- Therefore, the moles of nitrogen gas (N₂) needed for 19.11 moles of ammonia (NH₃) is:
[tex]\[
\text{moles of N₂} = \frac{\text{moles of NH₃}}{2} = \frac{19.11}{2} \approx 9.56 \, \text{moles}
\][/tex]
4. Determine the molar mass of nitrogen gas (N₂):
- Nitrogen gas (N₂) has a molar mass of:
[tex]\[
2 \times 14 \, \text{g/mol} = 28 \, g/mol
\][/tex]
5. Calculate the mass of nitrogen gas (N₂) required:
- Moles of nitrogen gas (N₂) = 9.56
- Molar mass of N₂ = 28 g/mol
- Mass of nitrogen gas needed is calculated using the formula:
[tex]\[
\text{mass of N₂} = \text{moles of N₂} \times \text{molar mass of N₂} = 9.56 \, \text{moles} \times 28 \, \text{g/mol} \approx 267.65 \, \text{g}
\][/tex]
Thus, the mass of nitrogen gas needed to produce 325 grams of ammonia is approximately 267.7 g. Therefore, the correct answer is:
(C) 267.7 g
