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17. Which of the following reaction mixtures would produce the greatest amount of product, assuming all went to completion? Each involves the reaction symbolized by the equation:

[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]

A. 2 mol [tex]\( N_2 \)[/tex] and 4 mol [tex]\( H_2 \)[/tex]

B. 1 mol [tex]\( N_2 \)[/tex] and 5 mol [tex]\( H_2 \)[/tex]

C. 4 mol [tex]\( N_2 \)[/tex] and 2 mol [tex]\( H_2 \)[/tex]

D. 5 mol [tex]\( N_2 \)[/tex] and 1 mol [tex]\( H_2 \)[/tex]



Answer :

GinnyAnswer
To determine which reaction mixture would produce the greatest amount of [tex]\( NH_3 \)[/tex] (ammonia), we need to carefully analyze the quantities of [tex]\( N_2 \)[/tex] (nitrogen) and [tex]\( H_2 \)[/tex] (hydrogen) given for each option and identify the limiting reactant in each case.

The balanced chemical equation for the reaction is:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]

1. Option (A): [tex]\(2 \, \text{mol} \, N_2\)[/tex] and [tex]\(4 \, \text{mol} \, H_2\)[/tex]
- To fully react [tex]\(4 \, \text{mol} \, H_2\)[/tex] (which is less than the [tex]\(6 \, \text{mol} \, H_2\)[/tex] needed for [tex]\(2 \, \text{mol} \, N_2\)[/tex]), we use:
[tex]\[ \frac{4 \, \text{mol} \, H_2}{3 \, \text{mol} \, H_2/\text{mol} \, N_2} = 1.333 \, \text{mol} \, N_2 \][/tex]
- The [tex]\(1.333 \, \text{mol} \, N_2\)[/tex] will react completely:
[tex]\[ 1.333 \, \text{mol} \, N_2 \times \frac{2 \, \text{mol} \, NH_3}{1 \, \text{mol} \, N_2} = 2.666 \, \text{mol} \, NH_3 \][/tex]

2. Option (B): [tex]\(1 \, \text{mol} \, N_2\)[/tex] and [tex]\(5 \, \text{mol} \, H_2\)[/tex]
- To fully react [tex]\(1 \, \text{mol} \, N_2\)[/tex], we use:
[tex]\[ 1 \, \text{mol} \, N_2 \times 3 \, \text{mol} \, H_2/\text{mol} \, N_2 = 3 \, \text{mol} \, H_2 \][/tex]
- Since [tex]\(5 \, \text{mol} \, H_2\)[/tex] is available, [tex]\(1 \, \text{mol} \, N_2\)[/tex] will completely react:
[tex]\[ 1 \, \text{mol} \, N_2 \times \frac{2 \, \text{mol} \, NH_3}{1 \, \text{mol} \, N_2} = 2 \, \text{mol} \, NH_3 \][/tex]

3. Option (C): [tex]\(4 \, \text{mol} \, N_2\)[/tex] and [tex]\(2 \, \text{mol} \, H_2\)[/tex]
- To fully react [tex]\(2 \, \text{mol} \, H_2\)[/tex] (which is less than the [tex]\(12 \, \text{mol} \, H_2\)[/tex] needed for [tex]\(4 \, \text{mol} \, N_2\)[/tex]), we use:
[tex]\[ \frac{2 \, \text{mol} \, H_2}{3 \, \text{mol} \, H_2/\text{mol} \, N_2} = 0.6667 \, \text{mol} \, N_2 \][/tex]
- The [tex]\(0.6667 \, \text{mol} \, N_2\)[/tex] will react completely:
[tex]\[ 0.6667 \, \text{mol} \, N_2 \times \frac{2 \, \text{mol} \, NH_3}{1 \, \text{mol} \, N_2} = 1.333 \, \text{mol} \, NH_3 \][/tex]

4. Option (D): [tex]\(5 \, \text{mol} \, N_2\)[/tex] and [tex]\(1 \, \text{mol} \, H_2\)[/tex]
- To fully react [tex]\(1 \, \text{mol} \, H_2\)[/tex] (which is less than the [tex]\(15 \, \text{mol} \, H_2\)[/tex] needed for [tex]\(5 \, \text{mol} \, N_2\)[/tex]), we use:
[tex]\[ \frac{1 \, \text{mol} \, H_2}{3 \, \text{mol} \, H_2/\text{mol} \, N_2} = 0.3333 \, \text{mol} \, N_2 \][/tex]
- The [tex]\(0.3333 \, \text{mol} \, N_2\)[/tex] will react completely:
[tex]\[ 0.3333 \, \text{mol} \, N_2 \times \frac{2 \, \text{mol} \, NH_3}{1 \, \text{mol} \, N_2} = 0.667 \, \text{mol} \, NH_3 \][/tex]

Comparing the results:

- Option (A) produces [tex]\(2.666 \, \text{mol} \, NH_3\)[/tex]
- Option (B) produces [tex]\(2 \, \text{mol} \, NH_3\)[/tex]
- Option (C) produces [tex]\(1.333 \, \text{mol} \, NH_3\)[/tex]
- Option (D) produces [tex]\(0.667 \, \text{mol} \, NH_3\)[/tex]

Thus, the greatest amount of [tex]\( NH_3 \)[/tex] produced is [tex]\(2.666 \, \text{mol} \, NH_3\)[/tex], corresponding to Option (A).

Therefore, the reaction mixture that would produce the greatest amount of product is:
[tex]\[ \boxed{2 \, \text{mol} \, N_2 \text{ and } 4 \, \text{mol} \, H_2} \][/tex]

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