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Multiple Choice
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
2 Comments
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.
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.
Calculate the moles of H₂ using the volume at STP: \( \text{moles of H}_2 = \frac{0.800 \text{ L}}{22.4 \text{ L/mol}} \).
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, \( \text{moles of N}_2 = \frac{\text{moles of H}_2}{3} \).
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: \( \text{mass of N}_2 = \text{moles of N}_2 \times 28.02 \text{ g/mol} \).