Question

how many liters of NH3 gas could be prepared by reacting 750L of nitrogen gas with an excess of hydrogen gas in the following reaction?
N2(g)+ 3H2(g)----->2 NH3(g)

Answer 1

Answer on Question #68506 - Chemistry - Other

Task:

How many liters of $\mathrm{NH}_3$ gas could be prepared by reacting 750L of nitrogen gas with an excess of hydrogen gas in the following reaction? $\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \longrightarrow 2\mathrm{NH}_3(\mathrm{g})$.

Solution:

Let us find the amount of nitrogen $(\mathrm{N}_2)$:

$$n(N_2) = \frac{V(N_2)}{Vm} = \frac{750L}{22.4 L/mol} = 33.482 \text{ moles of } N_2.$$

The reaction equation:

$$N_2(g) + 3H_2(g) = 2NH_3(g)$$

By the reaction equation: $n(N_2) = \frac{n(NH_3)}{2}$

Then,

$$\begin{array}{l} n(NH_3) = 2 \cdot n(N_2) = 2 \cdot 33.482 = 66.964 \text{ moles of } NH_3. \\ V(NH_3) = n(NH_3) \cdot Vm = 66.964 \text{ mol} \cdot 22.4 L/mol = 1500 L. \end{array}$$

Answer: 1500 liters of $\mathrm{NH}_3$.

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