Answer :
To solve this problem, we need to use the stoichiometric relationship given by the balanced chemical equation:
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]
This equation tells us that one mole of nitrogen gas ([tex]\(N_2\)[/tex]) reacts with three moles of hydrogen gas ([tex]\(H_2\)[/tex]) to produce two moles of ammonia ([tex]\(NH_3\)[/tex]).
Given:
- We have 3 moles of [tex]\(H_2\)[/tex].
We need to find:
- The moles of [tex]\(N_2\)[/tex] required to react with 3 moles of [tex]\(H_2\)[/tex].
### Solution:
1. Identify the Stoichiometric Ratio:
From the balanced chemical equation, the stoichiometric ratio of [tex]\(N_2\)[/tex] to [tex]\(H_2\)[/tex] is 1:3.
This means:
[tex]\[ \text{1 mole of } N_2 \text{ reacts with 3 moles of } H_2 \][/tex]
2. Set up the Relationship:
If we denote the moles of [tex]\(N_2\)[/tex] as [tex]\(x\)[/tex], then from the stoichiometric ratio:
[tex]\[ x \text{ moles of } N_2 \text{ react with } 3x \text{ moles of } H_2 \][/tex]
3. Solve for [tex]\(x\)[/tex]:
We are given that we have 3 moles of [tex]\(H_2\)[/tex]. Using the stoichiometric ratio:
[tex]\[ 3x = 3 \text{ moles of } H_2 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{3 \text{ moles of } H_2}{3} = 1 \text{ mole of } N_2 \][/tex]
Hence, 1 mole of [tex]\(N_2\)[/tex] will react with 3 moles of [tex]\(H_2\)[/tex].
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]
This equation tells us that one mole of nitrogen gas ([tex]\(N_2\)[/tex]) reacts with three moles of hydrogen gas ([tex]\(H_2\)[/tex]) to produce two moles of ammonia ([tex]\(NH_3\)[/tex]).
Given:
- We have 3 moles of [tex]\(H_2\)[/tex].
We need to find:
- The moles of [tex]\(N_2\)[/tex] required to react with 3 moles of [tex]\(H_2\)[/tex].
### Solution:
1. Identify the Stoichiometric Ratio:
From the balanced chemical equation, the stoichiometric ratio of [tex]\(N_2\)[/tex] to [tex]\(H_2\)[/tex] is 1:3.
This means:
[tex]\[ \text{1 mole of } N_2 \text{ reacts with 3 moles of } H_2 \][/tex]
2. Set up the Relationship:
If we denote the moles of [tex]\(N_2\)[/tex] as [tex]\(x\)[/tex], then from the stoichiometric ratio:
[tex]\[ x \text{ moles of } N_2 \text{ react with } 3x \text{ moles of } H_2 \][/tex]
3. Solve for [tex]\(x\)[/tex]:
We are given that we have 3 moles of [tex]\(H_2\)[/tex]. Using the stoichiometric ratio:
[tex]\[ 3x = 3 \text{ moles of } H_2 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{3 \text{ moles of } H_2}{3} = 1 \text{ mole of } N_2 \][/tex]
Hence, 1 mole of [tex]\(N_2\)[/tex] will react with 3 moles of [tex]\(H_2\)[/tex].
