Question

20. Ammonia is manufactured by the reaction of N2 and H2. An equilibrium mixture contains 5.0 g of
each N2, H2 and NH3. Calculate mass of N2 and H2 present initially.

Answer 1

Answer on the question #41591, Chemistry, Inorganic Chemistry

Question:

20. Ammonia is manufactured by the reaction of N2 and H2. An equilibrium mixture contains 5.0 g of each N2, H2 and NH3. Calculate mass of N2 and H2 present initially.

Solution:

$$N_2 + 3H_2 = 2NH_3$$

According to the reaction equation, the amounts of nitrogen, hydrogen and ammonia relate as:

$$n(N_2) = \frac{n(H_2)}{3} = \frac{n(NH_3)}{2}$$

Initial mass of nitrogen and hydrogen can be calculated as the sum of equilibrium mass and mass, that reacted with ammonia formation:

$$m(\text{initial}) = [m] + \frac{n(NH_3)}{i} * M$$

where $i$ – coefficient, that encounters stechiometric relations, $M$ – molar mass of the substance.

$$\begin{array}{l} m(N_2)_{ini} = [N_2] + \frac{n(NH_3)}{2} * M(N_2) = [N_2] + \frac{m(NH_3)}{2 * M(NH_3)} * M(N_2) = 5 + \frac{5}{2 * 17} * 28 \\ = 9.12 \text{ g} \\ \end{array}$$ $$m(H_2)_{ini} = [H_2] + m(NH_3) - \frac{n(NH_3)}{2} * M(N_2) = 5 + 5 - 4.12 = 5.88 \text{ g}$$

**Answer:** 9.12 and 5.88 g

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