What volume of ammonia can be obtained from 1.12 m3 of nitrogen if the reaction yield is 50% of the theoretically possible?

1.Let’s find the amount of substance nitrogen.

n = V: Vn

n = 1.12 m3: 22.4 m / kmol = 0.05 kmol.

N2 + 3H2 = 2NH3

For 1 mol of N2, there are 2 mol of NH3.

The substances are in quantitative ratios 1: 2. The amount of ammonia substance will be 2 times more than N2.

n (NH3) = 2n (N2) = 2 × 0.05 = 0.1 kmol.

2.Let’s find the volume of ammonia.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.1 kmol × 22.4 m3 / kmol = 2.24 m3.

3. 2.24 were calculated. According to the condition of the problem, the yield of ammonia is 50% of the theoretically possible.

2.24 m3 – 100%,

Vpractical – 50%,

Vpractical = (2.24 m3 × 50%): 100% = 1.12 m3.

Answer: 1.12 m3.