Question

9. What is the maximum amount of ammonia formed when 14 gm of N2 is mixed with 2 gm of H2.

Answer 1

Answer on Question #41594, Chemistry, Other

Question:

What is the maximum amount of ammonia formed when 14 gm of N2 is mixed with 2 gm of H2.

Solution:

Reaction equation is:

$$N _ {2} + 3 H _ {2} = 2 N H _ {3}$$

The maximum amount of ammonia can be calculated according to chemical equation:

$$n \left(N H _ {3}\right) = 2 n \left(N _ {2}\right), n \left(N H _ {3}\right) = 2 / 3 n \left(H _ {2}\right)$$ $$n \left(N _ {2}\right) = m \left(N _ {2}\right) / M \left(N _ {2}\right) = 1 4 / 2 8 = 0. 5 \mathrm {m o l}$$ $$n \left(H _ {2}\right) = m \left(H _ {2}\right) / M \left(H _ {2}\right) = 2 / 2 = 1 \mathrm {m o l}$$ $$3 n \left(N _ {2}\right) = n \left(H _ {2}\right)$$

As there is lack of hydrogen, the amount of ammonia corresponds to the hydrogen amount in such way:

$$n \left(N H _ {3}\right) = 2 / 3 ^ {*} n \left(H _ {2}\right) = 2 / 3 \mathrm {m o l}$$ $$m \left(N H _ {3}\right) = n \left(N H _ {3}\right) ^ {*} M \left(N H _ {3}\right) = 2 / 3 ^ {*} 1 7 = 1 1. 3 3 \mathrm {g}$$

Answer: 11.33 g

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