Answered

6.3

37 g of carbon (C) reacts with an excess of oxygen and produces 65 dm³ of carbon dioxide (CO₂) gas at STP. The balanced equation for the reaction is as follows:

[tex]\[ C(s) + O_2(g) \rightarrow CO_2(g) \][/tex]

6.4

22.5 dm³ of hydrogen (H₂) and 30 dm³ of nitrogen (N₂) are placed in a container and produce 12 dm³ of ammonia (NH₃). The balanced equation for the reaction is:

[tex]\[ 3H_2(g) + N_2(g) \rightarrow 2NH_3(g) \][/tex]

6.4.1

Calculate the enthalpy change for this reaction:

[tex]\[ \Delta H = ... \][/tex]



Answer :

GinnyAnswer
Absolutely, let’s break down the process of solving these chemistry problems step by step.

## Problem 6.3

### Given Data:
- Mass of Carbon, [tex]\( m_C = 37 \)[/tex] grams
- Molar Mass of Carbon, [tex]\( M_C = 12 \)[/tex] g/mol
- Volume of CO[tex]\(_2\)[/tex] produced, [tex]\( V_{CO_2} = 65 \)[/tex] dm[tex]\(^3\)[/tex]
- Molar volume of a gas at STP, [tex]\( V_m = 22.4 \)[/tex] dm[tex]\(^3\)[/tex]/mol

### Solution Steps:

1. Calculate the number of moles of Carbon reacted:
[tex]\[ n_C = \frac{m_C}{M_C} = \frac{37 \text{ g}}{12 \text{ g/mol}} = 3.0833 \text{ mol} \][/tex]

2. Using the balanced chemical equation:
[tex]\[ C(s) + O_2(g) \rightarrow CO_2(g) \][/tex]
The mole ratio of Carbon to CO[tex]\(_2\)[/tex] is 1:1, so moles of CO[tex]\(_2\)[/tex] produced, [tex]\( n_{CO_2} = n_C = 3.0833 \)[/tex] mol.

3. Calculate the theoretical volume of CO[tex]\(_2\)[/tex] produced at STP:
[tex]\[ \text{Theoretical Volume} = n_{CO_2} \times V_m = 3.0833 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} = 69.0672 \text{ dm}^3 \][/tex]

4. Calculate the efficiency of the reaction:
[tex]\[ \text{Efficiency} = \left( \frac{V_{CO_2}}{\text{Theoretical Volume}} \right) \times 100\% = \left( \frac{65 \text{ dm}^3}{69.0672 \text{ dm}^3} \right) \times 100\% \approx 94.11\% \][/tex]

### Conclusion:
The efficiency of the reaction is approximately 94.11%.

## Problem 6.4

### Given Data:
- Volume of H[tex]\(_2\)[/tex], [tex]\( V_{H_2} = 22.5 \)[/tex] dm[tex]\(^3\)[/tex]
- Volume of N[tex]\(_2\)[/tex], [tex]\( V_{N_2} = 30 \)[/tex] dm[tex]\(^3\)[/tex]
- Volume of NH[tex]\(_3\)[/tex] produced, [tex]\( V_{NH_3} = 12 \)[/tex] dm[tex]\(^3\)[/tex]
- Molar volume of gas, [tex]\( V_m = 22.4 \)[/tex] dm[tex]\(^3\)[/tex]/mol

### Solution Steps:

1. Calculate the number of moles of H[tex]\(_2\)[/tex], N[tex]\(_2\)[/tex], and NH[tex]\(_3\)[/tex]:
[tex]\[ n_{H_2} = \frac{V_{H_2}}{V_m} = \frac{22.5 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 1.0045 \text{ mol} \][/tex]
[tex]\[ n_{N_2} = \frac{V_{N_2}}{V_m} = \frac{30 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 1.3393 \text{ mol} \][/tex]
[tex]\[ n_{NH_3} = \frac{V_{NH_3}}{V_m} = \frac{12 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 0.5357 \text{ mol} \][/tex]

2. Using the balanced chemical equation:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]
The mole ratio of N[tex]\(_2\)[/tex] to NH[tex]\(_3\)[/tex] is 1:2, so:
[tex]\[ \text{Required moles of } N_2 = \frac{n_{NH_3}}{2} = \frac{0.5357 \text{ mol}}{2} = 0.2679 \text{ mol} \][/tex]

Similarly, the mole ratio of H[tex]\(_2\)[/tex] to NH[tex]\(_3\)[/tex] is 3:2, so:
[tex]\[ \text{Required moles of } H_2 = 3 \times 0.2679 = 0.8036 \text{ mol} \][/tex]

3. Calculate the percentage yield of NH[tex]\(_3\)[/tex]:
[tex]\[ \text{Percentage yield} = \left( \frac{n_{NH_3}}{\text{Required moles of } N_2} \right) \times 100\% = \left( \frac{0.5357 \text{ mol}}{0.2679 \text{ mol}} \right) \times 100\% = 200\% \][/tex]

### Conclusion:
- Required moles of H[tex]\(_2\)[/tex]: 0.8036 mol
- Required moles of N[tex]\(_2\)[/tex]: 0.2679 mol
- Percentage yield of NH[tex]\(_3\)[/tex]: 200%

Summarizing, the efficiency for the CO[tex]\(_2\)[/tex] production is approximately 94.11%, and the percentage yield of NH[tex]\(_3\)[/tex] is 200% with required reactant moles calculated as indicated.