Answer :
### Detailed Solution
#### Part 6.3: Calculation of Percentage Purity of Carbon
Given:
- Mass of carbon sample, [tex]\( m_{\text{C sample}} = 37 \)[/tex] g
- Volume of [tex]\( CO_2 \)[/tex] produced, [tex]\( V_{CO_2} = 65 \)[/tex] dm³
- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 dm³
- Molecular weight of carbon, [tex]\( M_{\text{C}} = 12 \)[/tex] g/mol
Reaction:
[tex]\[ C (s) + O_2 (g) \rightarrow CO_2 (g) \][/tex]
Steps:
1. Calculate the number of moles of [tex]\( CO_2 \)[/tex] produced:
[tex]\[ \text{Moles of } CO_2 = \frac{V_{CO_2}}{22.4 \text{ dm}^3/\text{mol}} = \frac{65 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} \approx 2.9018 \text{ moles} \][/tex]
2. The reaction shows a 1:1 molar ratio between carbon and [tex]\( CO_2 \)[/tex]. Therefore, the moles of carbon required are equal to the moles of [tex]\( CO_2 \)[/tex] produced:
[tex]\[ \text{Moles of } C = \text{Moles of } CO_2 = 2.9018 \text{ moles} \][/tex]
3. Calculate the mass of pure carbon required (using its molar mass):
[tex]\[ \text{Mass of pure carbon} = \text{Moles of } C \times M_{\text{C}} = 2.9018 \text{ moles} \times 12 \text{ g/mol} \approx 34.8214 \text{ g} \][/tex]
4. Calculate the percentage purity of the carbon sample:
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure carbon}}{\text{Mass of carbon sample}} \right) \times 100 = \left( \frac{34.8214 \text{ g}}{37 \text{ g}} \right) \times 100 \approx 94.11\% \][/tex]
#### Part 6.4: Calculation of Percentage Yield for the Reaction
Given:
- Volume of [tex]\( H_2 \)[/tex] placed in-container, [tex]\( V_{H_2} = 22.5 \)[/tex] dm³
- Volume of [tex]\( N_2 \)[/tex] placed in-container, [tex]\( V_{N_2} = 30 \)[/tex] dm³
- Volume of [tex]\( NH_3 \)[/tex] produced, [tex]\( V_{NH_3} = 12 \)[/tex] dm³
Reaction:
[tex]\[ N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g) \][/tex]
Steps:
1. Calculate the theoretical volume of [tex]\( NH_3 \)[/tex] that could be produced from [tex]\( H_2 \)[/tex]:
[tex]\[ \text{Theoretical } NH_3 \text{ from } H_2 = \frac{V_{H_2}}{3} \times 2 = \frac{22.5 \text{ dm}^3}{3} \times 2 = 15 \text{ dm}^3 \][/tex]
2. Calculate the theoretical volume of [tex]\( NH_3 \)[/tex] that could be produced from [tex]\( N_2 \)[/tex]:
[tex]\[ \text{Theoretical } NH_3 \text{ from } N_2 = V_{N_2} \times 2 = 30 \text{ dm}^3 \times 2 = 60 \text{ dm}^3 \][/tex]
3. Identify the limiting reagent and determine the theoretical yield:
Since the volume of [tex]\( NH_3 \)[/tex] from [tex]\( H_2 \)[/tex] (15 dm³) is less than that from [tex]\( N_2 \)[/tex] (60 dm³), [tex]\( H_2 \)[/tex] is the limiting reagent. Hence, the theoretical yield of [tex]\( NH_3 \)[/tex] is 15 dm³.
4. Calculate the percentage yield of [tex]\( NH_3 \)[/tex]:
[tex]\[ \text{Percentage yield} = \left( \frac{V_{NH_3 \text{ produced}}}{\text{Theoretical } NH_3} \right) \times 100 = \left( \frac{12 \text{ dm}^3}{15 \text{ dm}^3} \right) \times 100 \approx 80\% \][/tex]
### Conclusion
- The percentage purity of the carbon sample is approximately [tex]\( 94.11\% \)[/tex].
- The percentage yield for the reaction is [tex]\( 80\% \)[/tex].
#### Part 6.3: Calculation of Percentage Purity of Carbon
Given:
- Mass of carbon sample, [tex]\( m_{\text{C sample}} = 37 \)[/tex] g
- Volume of [tex]\( CO_2 \)[/tex] produced, [tex]\( V_{CO_2} = 65 \)[/tex] dm³
- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 dm³
- Molecular weight of carbon, [tex]\( M_{\text{C}} = 12 \)[/tex] g/mol
Reaction:
[tex]\[ C (s) + O_2 (g) \rightarrow CO_2 (g) \][/tex]
Steps:
1. Calculate the number of moles of [tex]\( CO_2 \)[/tex] produced:
[tex]\[ \text{Moles of } CO_2 = \frac{V_{CO_2}}{22.4 \text{ dm}^3/\text{mol}} = \frac{65 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} \approx 2.9018 \text{ moles} \][/tex]
2. The reaction shows a 1:1 molar ratio between carbon and [tex]\( CO_2 \)[/tex]. Therefore, the moles of carbon required are equal to the moles of [tex]\( CO_2 \)[/tex] produced:
[tex]\[ \text{Moles of } C = \text{Moles of } CO_2 = 2.9018 \text{ moles} \][/tex]
3. Calculate the mass of pure carbon required (using its molar mass):
[tex]\[ \text{Mass of pure carbon} = \text{Moles of } C \times M_{\text{C}} = 2.9018 \text{ moles} \times 12 \text{ g/mol} \approx 34.8214 \text{ g} \][/tex]
4. Calculate the percentage purity of the carbon sample:
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure carbon}}{\text{Mass of carbon sample}} \right) \times 100 = \left( \frac{34.8214 \text{ g}}{37 \text{ g}} \right) \times 100 \approx 94.11\% \][/tex]
#### Part 6.4: Calculation of Percentage Yield for the Reaction
Given:
- Volume of [tex]\( H_2 \)[/tex] placed in-container, [tex]\( V_{H_2} = 22.5 \)[/tex] dm³
- Volume of [tex]\( N_2 \)[/tex] placed in-container, [tex]\( V_{N_2} = 30 \)[/tex] dm³
- Volume of [tex]\( NH_3 \)[/tex] produced, [tex]\( V_{NH_3} = 12 \)[/tex] dm³
Reaction:
[tex]\[ N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g) \][/tex]
Steps:
1. Calculate the theoretical volume of [tex]\( NH_3 \)[/tex] that could be produced from [tex]\( H_2 \)[/tex]:
[tex]\[ \text{Theoretical } NH_3 \text{ from } H_2 = \frac{V_{H_2}}{3} \times 2 = \frac{22.5 \text{ dm}^3}{3} \times 2 = 15 \text{ dm}^3 \][/tex]
2. Calculate the theoretical volume of [tex]\( NH_3 \)[/tex] that could be produced from [tex]\( N_2 \)[/tex]:
[tex]\[ \text{Theoretical } NH_3 \text{ from } N_2 = V_{N_2} \times 2 = 30 \text{ dm}^3 \times 2 = 60 \text{ dm}^3 \][/tex]
3. Identify the limiting reagent and determine the theoretical yield:
Since the volume of [tex]\( NH_3 \)[/tex] from [tex]\( H_2 \)[/tex] (15 dm³) is less than that from [tex]\( N_2 \)[/tex] (60 dm³), [tex]\( H_2 \)[/tex] is the limiting reagent. Hence, the theoretical yield of [tex]\( NH_3 \)[/tex] is 15 dm³.
4. Calculate the percentage yield of [tex]\( NH_3 \)[/tex]:
[tex]\[ \text{Percentage yield} = \left( \frac{V_{NH_3 \text{ produced}}}{\text{Theoretical } NH_3} \right) \times 100 = \left( \frac{12 \text{ dm}^3}{15 \text{ dm}^3} \right) \times 100 \approx 80\% \][/tex]
### Conclusion
- The percentage purity of the carbon sample is approximately [tex]\( 94.11\% \)[/tex].
- The percentage yield for the reaction is [tex]\( 80\% \)[/tex].
