Question

Consider the following balanced equation: N2 + 3H2 --- 2NH3

When 4.50 mole of H2 react, how many grams of N2 are needed?

How many grams of H2 are required to produce 27.0 g of NH3?

Answer 1

Answer on Question #52136, Chemistry, Other

**Task:**

Consider the following balanced equation: $N_2 + 3H_2 \rightarrow 2NH_3$

a) When 4.50 mole of $H_2$ react, how many grams of $N_2$ are needed?

b) How many grams of $H_2$ are required to produce 27.0 g of $NH_3$?

**Answer:**

a) According to the given equation $N_2 + 3H_2 \rightarrow 2NH_3$, number of moles of $H_2$ is 3 times greater than $N_2$.

$$v(N_2) = \frac{v(H_2)}{3} = \frac{4.5}{3} = 1.5 \text{ moles}$$ $$v = \frac{m}{M} \quad m = vM$$ $$M(N_2) = 28 \text{ g/mol}$$ $$m(N_2) = 1.5 \cdot 28 = 42 \text{ g}$$ $$v(NH_3) = \frac{m(NH_3)}{M(NH_3)}$$ $$M(NH_3) = 17.0 \text{ g/mol}$$

b)

$$v(NH_3) = \frac{27.0}{17.0} = 1.6 \text{ mol}$$ $$v(H_2) = \frac{3 \cdot v(NH_3)}{2} = \frac{3 \cdot 1.6}{2} = 2.4 \text{ moles}$$ $$M(H_2) = 2 \text{ g/mol}$$ $$m(H_2) = 2.4 \cdot 2 = 4.8 \text{ g}$$

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