Nitrogen and hydrogen combine to form ammonia via the following reaction:

                      1 N₂ (s)  +  3 H₂ (g) → 2 NH₃ (g)

What mass of nitrogen is required to completely react with 800.0 mL H₂ at STP?

• A. 0.333 g
• B. 0.512 g
• C. 0.802 g
• D. 1.01 g
• E. 1.36 g

Answer: (A)

Verified step by step guidance

1. Start by understanding the reaction: 1 N₂ (s) + 3 H₂ (g) → 2 NH₃ (g). This tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
2. Since the problem states that hydrogen is at STP (Standard Temperature and Pressure), use the ideal gas law to find the number of moles of H₂. At STP, 1 mole of any gas occupies 22.4 L. Convert 800.0 mL of H₂ to liters by dividing by 1000, which gives 0.800 L.
3. Calculate the moles of H₂ using the volume at STP: \( \text{moles of H}_2 = \frac{0.800 \text{ L}}{22.4 \text{ L/mol}} \).
4. Use the stoichiometry of the balanced equation to find the moles of N₂ needed. According to the equation, 3 moles of H₂ react with 1 mole of N₂. Therefore, \( \text{moles of N}_2 = \frac{\text{moles of H}_2}{3} \).
5. Finally, calculate the mass of nitrogen required using its molar mass. The molar mass of N₂ is 28.02 g/mol. Multiply the moles of N₂ by the molar mass to find the mass: \( \text{mass of N}_2 = \text{moles of N}_2 \times 28.02 \text{ g/mol} \).