Question

1. ammonia is a principal nitrogen fertilizer. it is prepared by the reaction between hydrogen and nitrogen.

3 h₂(g) + n₂(g) → 2 nh₃(g)
5.0 moles of h₂ and 7.5 moles of n₂ were placed in a reaction vessel. how many moles of nh₃ can be produced? which substance is the limiting reactant? which substance is the excess reactant?

Explanation:

Step1: Determine the moles of $NH_3$ produced from each reactant

Based on the stoichiometry of the reaction $3H_2(g)+N_2(g)
ightarrow 2NH_3(g)$.
For $H_2$:
The mole - ratio of $H_2$ to $NH_3$ is $\frac{3}{2}$. Given $n_{H_2}=5.0$ moles. The moles of $NH_3$ produced from $H_2$, $n_{NH_3 - H_2}$ is calculated as follows:
$n_{NH_3 - H_2}=\frac{2}{3}\times n_{H_2}=\frac{2}{3}\times5.0=\frac{10}{3}\approx3.33$ moles.
For $N_2$:
The mole - ratio of $N_2$ to $NH_3$ is $\frac{1}{2}$. Given $n_{N_2}=7.5$ moles. The moles of $NH_3$ produced from $N_2$, $n_{NH_3 - N_2}$ is calculated as follows:
$n_{NH_3 - N_2}=2\times n_{N_2}=2\times7.5 = 15$ moles.

Step2: Identify the limiting and excess reactants

The limiting reactant is the one that produces the least amount of product. Since $\frac{10}{3}<15$, $H_2$ is the limiting reactant.
The excess reactant is the one that is not completely consumed. So, $N_2$ is the excess reactant.

Step3: Determine the amount of $NH_3$ produced

The amount of $NH_3$ produced is determined by the limiting reactant. So, the moles of $NH_3$ produced is $\frac{10}{3}\approx3.33$ moles.

Answer:

The moles of $NH_3$ produced is $\frac{10}{3}\approx3.33$ moles. The limiting reactant is $H_2$. The excess reactant is $N_2$.