How many grams of N_2 are required to produce 100.0 L of NH_3 at STP according to the reaction N_2 + 3H_2 → 2NH_3?

• A. 41.8 g
• B. 56.0 g
• C. 28.0 g
• D. 83.6 g

Answer: (D)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}\).
2. Identify the volume of ammonia (\(\mathrm{NH_3}\)) given: 100.0 L at STP. Recall that at STP, 1 mole of any gas occupies 22.4 L.
3. Calculate the moles of \(\mathrm{NH_3}\) produced using the formula: \(\text{moles of } NH_3 = \frac{\text{volume of } NH_3}{22.4 \text{ L/mol}}\).
4. Use the mole ratio from the balanced equation to find moles of \(\mathrm{N_2}\) required. According to the equation, 2 moles of \(\mathrm{NH_3}\) are produced per 1 mole of \(\mathrm{N_2}\), so: \(\text{moles of } N_2 = \frac{1}{2} \times \text{moles of } NH_3\).
5. Convert moles of \(\mathrm{N_2}\) to grams using the molar mass of nitrogen gas (\(\mathrm{N_2}\)), which is approximately 28.0 g/mol: \(\text{mass of } N_2 = \text{moles of } N_2 \times 28.0 \text{ g/mol}\).