Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
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Isolated i believe dont hold me to that
