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Answer :
To solve the problem of how much heat is released when 50.0 g of ammonia is condensed at [tex]\(-33^{\circ} C\)[/tex], we can follow these steps:
1. Determine the Molar Mass of Ammonia ([tex]\(NH_3\)[/tex]):
The chemical formula [tex]\(NH_3\)[/tex] consists of:
- Nitrogen (N): approximately 14.01 g/mol
- Hydrogen (H): 1.008 g/mol, and since there are three hydrogens, we have [tex]\(3 \times 1.008 = 3.024\)[/tex] g/mol
Adding these together, the molar mass of ammonia is approximately [tex]\(14.01 + 3.024 = 17.034\)[/tex] g/mol.
2. Calculate the Moles of Ammonia:
To find out how many moles are in 50.0 grams of ammonia, we use the formula:
[tex]\[
\text{moles of ammonia} = \frac{\text{mass of ammonia}}{\text{molar mass of ammonia}} = \frac{50.0 \text{ g}}{17.034 \text{ g/mol}} \approx 2.94 \text{ moles}
\][/tex]
3. Utilize the Enthalpy of Vaporization:
The standard enthalpy of vaporization for ammonia is [tex]\(23.3 \text{ kJ/mol}\)[/tex], but since we are considering the condensation process (which is the reverse of vaporization), the process is exothermic, meaning heat is released, so it is negative.
4. Calculate the Heat Released:
The heat released when ammonia is condensed can be calculated by:
[tex]\[
\text{heat released} = -(\text{moles of ammonia}) \times (\text{enthalpy of vaporization})
\][/tex]
Substituting the values:
[tex]\[
\text{heat released} = -(2.94 \text{ moles}) \times 23.3 \text{ kJ/mol} \approx -68.4 \text{ kJ}
\][/tex]
So, the correct answer is [tex]\( \boxed{-68.4 \, \text{kJ}} \)[/tex], which corresponds to option A.
1. Determine the Molar Mass of Ammonia ([tex]\(NH_3\)[/tex]):
The chemical formula [tex]\(NH_3\)[/tex] consists of:
- Nitrogen (N): approximately 14.01 g/mol
- Hydrogen (H): 1.008 g/mol, and since there are three hydrogens, we have [tex]\(3 \times 1.008 = 3.024\)[/tex] g/mol
Adding these together, the molar mass of ammonia is approximately [tex]\(14.01 + 3.024 = 17.034\)[/tex] g/mol.
2. Calculate the Moles of Ammonia:
To find out how many moles are in 50.0 grams of ammonia, we use the formula:
[tex]\[
\text{moles of ammonia} = \frac{\text{mass of ammonia}}{\text{molar mass of ammonia}} = \frac{50.0 \text{ g}}{17.034 \text{ g/mol}} \approx 2.94 \text{ moles}
\][/tex]
3. Utilize the Enthalpy of Vaporization:
The standard enthalpy of vaporization for ammonia is [tex]\(23.3 \text{ kJ/mol}\)[/tex], but since we are considering the condensation process (which is the reverse of vaporization), the process is exothermic, meaning heat is released, so it is negative.
4. Calculate the Heat Released:
The heat released when ammonia is condensed can be calculated by:
[tex]\[
\text{heat released} = -(\text{moles of ammonia}) \times (\text{enthalpy of vaporization})
\][/tex]
Substituting the values:
[tex]\[
\text{heat released} = -(2.94 \text{ moles}) \times 23.3 \text{ kJ/mol} \approx -68.4 \text{ kJ}
\][/tex]
So, the correct answer is [tex]\( \boxed{-68.4 \, \text{kJ}} \)[/tex], which corresponds to option A.
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