Answer :
Certainly! Let's solve each part of the question step-by-step:
### 2.1 Mass of [tex]\( \text{NO} \)[/tex] Produced from 2 mol of [tex]\( \text{NH}_3 \)[/tex] with Excess [tex]\( \text{O}_2 \)[/tex]
1. Balanced Chemical Equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 7 \text{NO} \][/tex]
2. Determine the Stoichiometric Relationship:
From the equation, 4 moles of [tex]\( \text{NH}_3 \)[/tex] produce 7 moles of [tex]\( \text{NO} \)[/tex].
3. Mass Calculation:
- Given: 2 moles of [tex]\( \text{NH}_3 \)[/tex] react.
- Using stoichiometry,
[tex]\[ \text{moles of NO} = \left( \frac{2 \text{ moles NH}_3 \times 7 \text{ moles NO}}{4 \text{ moles NH}_3} \right) = 3.5 \text{ moles NO} \][/tex]
- Molar Mass of [tex]\( \text{NO} \)[/tex] is 30.01 g/mol.
- Mass of [tex]\( \text{NO} \)[/tex] produced:
[tex]\[ \text{mass of NO} = 3.5 \text{ moles NO} \times 30.01 \text{ g/mol} = 105.035 \text{ g} \][/tex]
So, 105.035 grams of [tex]\( \text{NO} \)[/tex] is produced.
### 2.2 Mass of [tex]\( \text{H}_2\text{O} \)[/tex] Produced from 4 mol of [tex]\( \text{O}_2 \)[/tex] with Excess [tex]\( \text{NH}_3 \)[/tex]
1. Balanced Chemical Equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 7 \text{NO} \][/tex]
2. Determine the Stoichiometric Relationship:
From the equation, 5 moles of [tex]\( \text{O}_2 \)[/tex] produce 6 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
3. Mass Calculation:
- Given: 4 moles of [tex]\( \text{O}_2 \)[/tex] react.
- Using stoichiometry,
[tex]\[ \text{moles of H}_2\text{O} = \left( \frac{4 \text{ moles O}_2 \times 6 \text{ moles H}_2\text{O}}{5 \text{ moles O}_2} \right) = 4.8 \text{ moles H}_2\text{O} \][/tex]
- Molar Mass of [tex]\( \text{H}_2\text{O} \)[/tex] is 18.015 g/mol.
- Mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced:
[tex]\[ \text{mass of H}_2\text{O} = 4.8 \text{ moles H}_2\text{O} \times 18.015 \text{ g/mol} = 86.472 \text{ g} \][/tex]
So, 86.472 grams of [tex]\( \text{H}_2\text{O} \)[/tex] is produced.
### 2.3 Mass of [tex]\( \text{O}_2 \)[/tex] Required to Form 2.5 mol of [tex]\( \text{NO} \)[/tex]
1. Balanced Chemical Equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 7 \text{NO} \][/tex]
2. Determine the Stoichiometric Relationship:
From the equation, 7 moles of [tex]\( \text{NO} \)[/tex] require 5 moles of [tex]\( \text{O}_2 \)[/tex].
3. Mass Calculation:
- Given: 2.5 moles of [tex]\( \text{NO} \)[/tex] to be formed.
- Using stoichiometry,
[tex]\[ \text{moles of O}_2 = \left( \frac{2.5 \text{ moles NO} \times 5 \text{ moles O}_2}{7 \text{ moles NO}} \right) = 1.785714 \text{ moles O}_2 \][/tex]
- Molar Mass of [tex]\( \text{O}_2 \)[/tex] is 32.00 g/mol.
- Mass of [tex]\( \text{O}_2 \)[/tex] required:
[tex]\[ \text{mass of O}_2 = 1.785714 \text{ moles O}_2 \times 32.00 \text{ g/mol} = 57.142857 \text{ g} \][/tex]
So, 57.142857 grams of [tex]\( \text{O}_2 \)[/tex] is required.
In summary:
1. 105.035 grams of [tex]\( \text{NO} \)[/tex] is produced from 2 moles of [tex]\( \text{NH}_3 \)[/tex].
2. 86.472 grams of [tex]\( \text{H}_2\text{O} \)[/tex] is produced from 4 moles of [tex]\( \text{O}_2 \)[/tex].
3. 57.142857 grams of [tex]\( \text{O}_2 \)[/tex] is required to form 2.5 moles of [tex]\( \text{NO} \)[/tex].
### 2.1 Mass of [tex]\( \text{NO} \)[/tex] Produced from 2 mol of [tex]\( \text{NH}_3 \)[/tex] with Excess [tex]\( \text{O}_2 \)[/tex]
1. Balanced Chemical Equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 7 \text{NO} \][/tex]
2. Determine the Stoichiometric Relationship:
From the equation, 4 moles of [tex]\( \text{NH}_3 \)[/tex] produce 7 moles of [tex]\( \text{NO} \)[/tex].
3. Mass Calculation:
- Given: 2 moles of [tex]\( \text{NH}_3 \)[/tex] react.
- Using stoichiometry,
[tex]\[ \text{moles of NO} = \left( \frac{2 \text{ moles NH}_3 \times 7 \text{ moles NO}}{4 \text{ moles NH}_3} \right) = 3.5 \text{ moles NO} \][/tex]
- Molar Mass of [tex]\( \text{NO} \)[/tex] is 30.01 g/mol.
- Mass of [tex]\( \text{NO} \)[/tex] produced:
[tex]\[ \text{mass of NO} = 3.5 \text{ moles NO} \times 30.01 \text{ g/mol} = 105.035 \text{ g} \][/tex]
So, 105.035 grams of [tex]\( \text{NO} \)[/tex] is produced.
### 2.2 Mass of [tex]\( \text{H}_2\text{O} \)[/tex] Produced from 4 mol of [tex]\( \text{O}_2 \)[/tex] with Excess [tex]\( \text{NH}_3 \)[/tex]
1. Balanced Chemical Equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 7 \text{NO} \][/tex]
2. Determine the Stoichiometric Relationship:
From the equation, 5 moles of [tex]\( \text{O}_2 \)[/tex] produce 6 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
3. Mass Calculation:
- Given: 4 moles of [tex]\( \text{O}_2 \)[/tex] react.
- Using stoichiometry,
[tex]\[ \text{moles of H}_2\text{O} = \left( \frac{4 \text{ moles O}_2 \times 6 \text{ moles H}_2\text{O}}{5 \text{ moles O}_2} \right) = 4.8 \text{ moles H}_2\text{O} \][/tex]
- Molar Mass of [tex]\( \text{H}_2\text{O} \)[/tex] is 18.015 g/mol.
- Mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced:
[tex]\[ \text{mass of H}_2\text{O} = 4.8 \text{ moles H}_2\text{O} \times 18.015 \text{ g/mol} = 86.472 \text{ g} \][/tex]
So, 86.472 grams of [tex]\( \text{H}_2\text{O} \)[/tex] is produced.
### 2.3 Mass of [tex]\( \text{O}_2 \)[/tex] Required to Form 2.5 mol of [tex]\( \text{NO} \)[/tex]
1. Balanced Chemical Equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 7 \text{NO} \][/tex]
2. Determine the Stoichiometric Relationship:
From the equation, 7 moles of [tex]\( \text{NO} \)[/tex] require 5 moles of [tex]\( \text{O}_2 \)[/tex].
3. Mass Calculation:
- Given: 2.5 moles of [tex]\( \text{NO} \)[/tex] to be formed.
- Using stoichiometry,
[tex]\[ \text{moles of O}_2 = \left( \frac{2.5 \text{ moles NO} \times 5 \text{ moles O}_2}{7 \text{ moles NO}} \right) = 1.785714 \text{ moles O}_2 \][/tex]
- Molar Mass of [tex]\( \text{O}_2 \)[/tex] is 32.00 g/mol.
- Mass of [tex]\( \text{O}_2 \)[/tex] required:
[tex]\[ \text{mass of O}_2 = 1.785714 \text{ moles O}_2 \times 32.00 \text{ g/mol} = 57.142857 \text{ g} \][/tex]
So, 57.142857 grams of [tex]\( \text{O}_2 \)[/tex] is required.
In summary:
1. 105.035 grams of [tex]\( \text{NO} \)[/tex] is produced from 2 moles of [tex]\( \text{NH}_3 \)[/tex].
2. 86.472 grams of [tex]\( \text{H}_2\text{O} \)[/tex] is produced from 4 moles of [tex]\( \text{O}_2 \)[/tex].
3. 57.142857 grams of [tex]\( \text{O}_2 \)[/tex] is required to form 2.5 moles of [tex]\( \text{NO} \)[/tex].
