Answer :
Let's solve each part of the question step-by-step.
### 1. Consider the balanced equation: [tex]\( N_2 + 3H_2 \rightarrow 2NH_3 \)[/tex]
#### 1.1. How many moles of [tex]\( N_2 \)[/tex] will react with 3 moles of [tex]\( H_2 \)[/tex]?
According to the balanced equation, 1 mole of [tex]\( N_2 \)[/tex] reacts with 3 moles of [tex]\( H_2 \)[/tex].
So, to react with 3 moles of [tex]\( H_2 \)[/tex]:
[tex]\[ N_2 \text{ moles} = \frac{3 \, \text{moles of} \, H_2}{3} = 1 \text{ mole of } N_2 \][/tex]
So, 1 mole of [tex]\( N_2 \)[/tex] will react with 3 moles of [tex]\( H_2 \)[/tex].
#### 1.2. How many moles of [tex]\( H_2 \)[/tex] will react with 3 moles of [tex]\( N_2 \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( H_2 \)[/tex] are required to react with 1 mole of [tex]\( N_2 \)[/tex].
So, to react with 3 moles of [tex]\( N_2 \)[/tex]:
[tex]\[ H_2 \text{ moles} = 3 \, (\text{moles of } N_2) \times 3 = 9 \text{ moles of } H_2 \][/tex]
So, 9 moles of [tex]\( H_2 \)[/tex] will react with 3 moles of [tex]\( N_2 \)[/tex].
#### 1.3. How many moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( H_2 \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( H_2 \)[/tex] produce 2 moles of [tex]\( NH_3 \)[/tex].
So, from 11 moles of [tex]\( H_2 \)[/tex]:
[tex]\[ NH_3 \text{ moles} = \frac{11 \, (\text{moles of } H_2) \times 2}{3} = 7.33 \, \text{(approximately)} \][/tex]
So, 7.33 moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( H_2 \)[/tex].
#### 1.4. How many moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( N_2 \)[/tex]?
According to the balanced equation, 1 mole of [tex]\( N_2 \)[/tex] produces 2 moles of [tex]\( NH_3 \)[/tex].
So, from 11 moles of [tex]\( N_2 \)[/tex]:
[tex]\[ NH_3 \text{ moles} = 11 \, (\text{moles of } N_2) \times 2 = 22 \text{ moles} \][/tex]
So, 22 moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( N_2 \)[/tex].
### 2. Consider the balanced equation: [tex]\( 3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2 \)[/tex]
#### 2.1. How many moles of [tex]\( Fe_3O_4 \)[/tex] will be formed from 12 moles of [tex]\( Fe \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( Fe \)[/tex] produce 1 mole of [tex]\( Fe_3O_4 \)[/tex].
So, from 12 moles of [tex]\( Fe \)[/tex]:
[tex]\[ Fe_3O_4 \text{ moles} = \frac{12 \, (\text{moles of } Fe)}{3} = 4 \text{ moles} \][/tex]
So, 4 moles of [tex]\( Fe_3O_4 \)[/tex] will be formed from 12 moles of [tex]\( Fe \)[/tex].
#### 2.2. How many moles of [tex]\( Fe \)[/tex] are needed to produce 16 moles of [tex]\( H_2 \)[/tex]?
According to the balanced equation, 4 moles of [tex]\( H_2 \)[/tex] are produced from 3 moles of [tex]\( Fe \)[/tex].
So, to produce 16 moles of [tex]\( H_2 \)[/tex]:
[tex]\[ Fe \text{ moles} = \frac{16 \, (\text{moles of } H_2) \times 3}{4} = 12 \text{ moles} \][/tex]
So, 12 moles of [tex]\( Fe \)[/tex] are needed to produce 16 moles of [tex]\( H_2 \)[/tex].
#### 2.3. How many moles of [tex]\( H_2 \)[/tex] will be formed if 40 moles of [tex]\( Fe_3O_4 \)[/tex] are formed?
According to the balanced equation, 1 mole of [tex]\( Fe_3O_4 \)[/tex] produces 4 moles of [tex]\( H_2 \)[/tex].
So, from 40 moles of [tex]\( Fe_3O_4 \)[/tex]:
[tex]\[ H_2 \text{ moles} = 40 \, (\text{moles of } Fe_3O_4) \times 4 = 160 \text{ moles} \][/tex]
So, 160 moles of [tex]\( H_2 \)[/tex] will be formed from 40 moles of [tex]\( Fe_3O_4 \)[/tex].
#### 2.4. How many moles of [tex]\( H_2O \)[/tex] are needed to react with 14.5 moles of [tex]\( Fe \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( Fe \)[/tex] react with 4 moles of [tex]\( H_2O \)[/tex].
So, to react with 14.5 moles of [tex]\( Fe \)[/tex]:
[tex]\[ H_2O \text{ moles} = \frac{14.5 \, (\text{moles of } Fe) \times 4}{3} = 19.33 \, \text{(approximately)} \][/tex]
So, 19.33 moles of [tex]\( H_2O \)[/tex] are needed to react with 14.5 moles of [tex]\( Fe \)[/tex].
### 1. Consider the balanced equation: [tex]\( N_2 + 3H_2 \rightarrow 2NH_3 \)[/tex]
#### 1.1. How many moles of [tex]\( N_2 \)[/tex] will react with 3 moles of [tex]\( H_2 \)[/tex]?
According to the balanced equation, 1 mole of [tex]\( N_2 \)[/tex] reacts with 3 moles of [tex]\( H_2 \)[/tex].
So, to react with 3 moles of [tex]\( H_2 \)[/tex]:
[tex]\[ N_2 \text{ moles} = \frac{3 \, \text{moles of} \, H_2}{3} = 1 \text{ mole of } N_2 \][/tex]
So, 1 mole of [tex]\( N_2 \)[/tex] will react with 3 moles of [tex]\( H_2 \)[/tex].
#### 1.2. How many moles of [tex]\( H_2 \)[/tex] will react with 3 moles of [tex]\( N_2 \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( H_2 \)[/tex] are required to react with 1 mole of [tex]\( N_2 \)[/tex].
So, to react with 3 moles of [tex]\( N_2 \)[/tex]:
[tex]\[ H_2 \text{ moles} = 3 \, (\text{moles of } N_2) \times 3 = 9 \text{ moles of } H_2 \][/tex]
So, 9 moles of [tex]\( H_2 \)[/tex] will react with 3 moles of [tex]\( N_2 \)[/tex].
#### 1.3. How many moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( H_2 \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( H_2 \)[/tex] produce 2 moles of [tex]\( NH_3 \)[/tex].
So, from 11 moles of [tex]\( H_2 \)[/tex]:
[tex]\[ NH_3 \text{ moles} = \frac{11 \, (\text{moles of } H_2) \times 2}{3} = 7.33 \, \text{(approximately)} \][/tex]
So, 7.33 moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( H_2 \)[/tex].
#### 1.4. How many moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( N_2 \)[/tex]?
According to the balanced equation, 1 mole of [tex]\( N_2 \)[/tex] produces 2 moles of [tex]\( NH_3 \)[/tex].
So, from 11 moles of [tex]\( N_2 \)[/tex]:
[tex]\[ NH_3 \text{ moles} = 11 \, (\text{moles of } N_2) \times 2 = 22 \text{ moles} \][/tex]
So, 22 moles of [tex]\( NH_3 \)[/tex] will be formed from 11 moles of [tex]\( N_2 \)[/tex].
### 2. Consider the balanced equation: [tex]\( 3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2 \)[/tex]
#### 2.1. How many moles of [tex]\( Fe_3O_4 \)[/tex] will be formed from 12 moles of [tex]\( Fe \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( Fe \)[/tex] produce 1 mole of [tex]\( Fe_3O_4 \)[/tex].
So, from 12 moles of [tex]\( Fe \)[/tex]:
[tex]\[ Fe_3O_4 \text{ moles} = \frac{12 \, (\text{moles of } Fe)}{3} = 4 \text{ moles} \][/tex]
So, 4 moles of [tex]\( Fe_3O_4 \)[/tex] will be formed from 12 moles of [tex]\( Fe \)[/tex].
#### 2.2. How many moles of [tex]\( Fe \)[/tex] are needed to produce 16 moles of [tex]\( H_2 \)[/tex]?
According to the balanced equation, 4 moles of [tex]\( H_2 \)[/tex] are produced from 3 moles of [tex]\( Fe \)[/tex].
So, to produce 16 moles of [tex]\( H_2 \)[/tex]:
[tex]\[ Fe \text{ moles} = \frac{16 \, (\text{moles of } H_2) \times 3}{4} = 12 \text{ moles} \][/tex]
So, 12 moles of [tex]\( Fe \)[/tex] are needed to produce 16 moles of [tex]\( H_2 \)[/tex].
#### 2.3. How many moles of [tex]\( H_2 \)[/tex] will be formed if 40 moles of [tex]\( Fe_3O_4 \)[/tex] are formed?
According to the balanced equation, 1 mole of [tex]\( Fe_3O_4 \)[/tex] produces 4 moles of [tex]\( H_2 \)[/tex].
So, from 40 moles of [tex]\( Fe_3O_4 \)[/tex]:
[tex]\[ H_2 \text{ moles} = 40 \, (\text{moles of } Fe_3O_4) \times 4 = 160 \text{ moles} \][/tex]
So, 160 moles of [tex]\( H_2 \)[/tex] will be formed from 40 moles of [tex]\( Fe_3O_4 \)[/tex].
#### 2.4. How many moles of [tex]\( H_2O \)[/tex] are needed to react with 14.5 moles of [tex]\( Fe \)[/tex]?
According to the balanced equation, 3 moles of [tex]\( Fe \)[/tex] react with 4 moles of [tex]\( H_2O \)[/tex].
So, to react with 14.5 moles of [tex]\( Fe \)[/tex]:
[tex]\[ H_2O \text{ moles} = \frac{14.5 \, (\text{moles of } Fe) \times 4}{3} = 19.33 \, \text{(approximately)} \][/tex]
So, 19.33 moles of [tex]\( H_2O \)[/tex] are needed to react with 14.5 moles of [tex]\( Fe \)[/tex].
