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### 9.1.1 Define the term 1 mole.
A mole is a unit that measures the amount of a substance. One mole of any substance contains exactly [tex]\(6.02214076 \times 10^{23}\)[/tex] of elementary entities (atoms, molecules, ions, electrons, etc). This constant number is known as Avogadro's number.
### 9.1.2 How many moles of ammonia will be produced from 1 mole of hydrogen gas?
The balanced chemical equation for the reaction is:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
From the equation, 3 moles of [tex]\(H_2\)[/tex] react to form 2 moles of [tex]\(NH_3\)[/tex].
To find out how many moles of [tex]\(NH_3\)[/tex] will be produced from 1 mole of [tex]\(H_2\)[/tex], we use the stoichiometric ratio from the balanced equation. This ratio is [tex]\(\frac{2}{3}\)[/tex].
Thus, for 1 mole of [tex]\(H_2\)[/tex]:
[tex]\[ \text{Moles of } NH_3 = 1 \text{ mole } H_2 \times \frac{2}{3} = 0.6666666666666666 \text{ moles of } NH_3 \][/tex]
### 9.1.3 Calculate the volume of gas that will remain in the container after the reaction is completed.
We start with 10 [tex]\(cm^3\)[/tex] of nitrogen ([tex]\(N_2\)[/tex]) and 24 [tex]\(cm^3\)[/tex] of hydrogen ([tex]\(H_2\)[/tex]). The balanced equation again is:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
According to the equation, 1 volume of [tex]\(N_2\)[/tex] reacts with 3 volumes of [tex]\(H_2\)[/tex] to produce 2 volumes of [tex]\(NH_3\)[/tex].
To determine which reactant is limiting, we use the initial volumes:
- [tex]\( N_2 \)[/tex]: 10 [tex]\(cm^3\)[/tex]
- [tex]\( H_2 \)[/tex]: 24 [tex]\(cm^3\)[/tex]
For the given volumes, to completely react with 10 [tex]\(cm^3\)[/tex] of [tex]\(N_2\)[/tex]:
[tex]\[ \text{Required } H_2 = 10 \ cm^3 \times 3 = 30 \ cm^3 \][/tex]
Since only 24 [tex]\(cm^3\)[/tex] of [tex]\(H_2\)[/tex] is available, [tex]\(H_2\)[/tex] is the limiting reactant.
### Volume of remaining gases
Because [tex]\(H_2\)[/tex] is the limiting reactant:
- [tex]\(H_2\)[/tex] will be used up completely.
- [tex]\(N_2\)[/tex] will be left partially unreacted.
The reaction uses [tex]\(H_2\)[/tex] entirely:
[tex]\[ 24 \ cm^3 \ H_2 \rightarrow \frac{24 \ cm^3 \ H_2}{3} = 8 \ cm^3 \ N_2 \][/tex]
Thus, remaining [tex]\(N_2\)[/tex]:
[tex]\[ 10 \ cm^3 - 8 \ cm^3 = 2 \ cm^3 \][/tex]
Volume of [tex]\(NH_3\)[/tex] produced:
[tex]\[ 2 \ volumes \ of \ NH_3 \ are \ produced \ for \ 3 \ volumes \ of \ H_2 \][/tex]
[tex]\[ 16 \ cm^3 \ NH_3 \left( \frac{2}{3} \times 24 \ cm^3 \right) \][/tex]
So, the total remaining gas volume, excluding ammonia:
[tex]\[ 2 \ cm^3 \ of \ N_2 + 0 \ cm^3 \ of \ H_2 = 2 \ cm^3 \][/tex]
Therefore, after the reaction is completed:
1. The moles of ammonia produced from 1 mole of hydrogen gas is 0.6666666666666666 moles.
2. The remaining volume of nitrogen gas in the container is 2 [tex]\(cm^3\)[/tex].
3. No remaining [tex]\(H_2\)[/tex] gas.
4. The volume of [tex]\(NH_3\)[/tex] produced is 16 [tex]\(cm^3\)[/tex].
5. The total remaining volume of gas excluding [tex]\(NH_3\)[/tex] is 2 [tex]\(cm^3\)[/tex].
### 9.1.1 Define the term 1 mole.
A mole is a unit that measures the amount of a substance. One mole of any substance contains exactly [tex]\(6.02214076 \times 10^{23}\)[/tex] of elementary entities (atoms, molecules, ions, electrons, etc). This constant number is known as Avogadro's number.
### 9.1.2 How many moles of ammonia will be produced from 1 mole of hydrogen gas?
The balanced chemical equation for the reaction is:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
From the equation, 3 moles of [tex]\(H_2\)[/tex] react to form 2 moles of [tex]\(NH_3\)[/tex].
To find out how many moles of [tex]\(NH_3\)[/tex] will be produced from 1 mole of [tex]\(H_2\)[/tex], we use the stoichiometric ratio from the balanced equation. This ratio is [tex]\(\frac{2}{3}\)[/tex].
Thus, for 1 mole of [tex]\(H_2\)[/tex]:
[tex]\[ \text{Moles of } NH_3 = 1 \text{ mole } H_2 \times \frac{2}{3} = 0.6666666666666666 \text{ moles of } NH_3 \][/tex]
### 9.1.3 Calculate the volume of gas that will remain in the container after the reaction is completed.
We start with 10 [tex]\(cm^3\)[/tex] of nitrogen ([tex]\(N_2\)[/tex]) and 24 [tex]\(cm^3\)[/tex] of hydrogen ([tex]\(H_2\)[/tex]). The balanced equation again is:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
According to the equation, 1 volume of [tex]\(N_2\)[/tex] reacts with 3 volumes of [tex]\(H_2\)[/tex] to produce 2 volumes of [tex]\(NH_3\)[/tex].
To determine which reactant is limiting, we use the initial volumes:
- [tex]\( N_2 \)[/tex]: 10 [tex]\(cm^3\)[/tex]
- [tex]\( H_2 \)[/tex]: 24 [tex]\(cm^3\)[/tex]
For the given volumes, to completely react with 10 [tex]\(cm^3\)[/tex] of [tex]\(N_2\)[/tex]:
[tex]\[ \text{Required } H_2 = 10 \ cm^3 \times 3 = 30 \ cm^3 \][/tex]
Since only 24 [tex]\(cm^3\)[/tex] of [tex]\(H_2\)[/tex] is available, [tex]\(H_2\)[/tex] is the limiting reactant.
### Volume of remaining gases
Because [tex]\(H_2\)[/tex] is the limiting reactant:
- [tex]\(H_2\)[/tex] will be used up completely.
- [tex]\(N_2\)[/tex] will be left partially unreacted.
The reaction uses [tex]\(H_2\)[/tex] entirely:
[tex]\[ 24 \ cm^3 \ H_2 \rightarrow \frac{24 \ cm^3 \ H_2}{3} = 8 \ cm^3 \ N_2 \][/tex]
Thus, remaining [tex]\(N_2\)[/tex]:
[tex]\[ 10 \ cm^3 - 8 \ cm^3 = 2 \ cm^3 \][/tex]
Volume of [tex]\(NH_3\)[/tex] produced:
[tex]\[ 2 \ volumes \ of \ NH_3 \ are \ produced \ for \ 3 \ volumes \ of \ H_2 \][/tex]
[tex]\[ 16 \ cm^3 \ NH_3 \left( \frac{2}{3} \times 24 \ cm^3 \right) \][/tex]
So, the total remaining gas volume, excluding ammonia:
[tex]\[ 2 \ cm^3 \ of \ N_2 + 0 \ cm^3 \ of \ H_2 = 2 \ cm^3 \][/tex]
Therefore, after the reaction is completed:
1. The moles of ammonia produced from 1 mole of hydrogen gas is 0.6666666666666666 moles.
2. The remaining volume of nitrogen gas in the container is 2 [tex]\(cm^3\)[/tex].
3. No remaining [tex]\(H_2\)[/tex] gas.
4. The volume of [tex]\(NH_3\)[/tex] produced is 16 [tex]\(cm^3\)[/tex].
5. The total remaining volume of gas excluding [tex]\(NH_3\)[/tex] is 2 [tex]\(cm^3\)[/tex].
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