To determine the quantity of heat released when 27.9 g of H2O condenses, we'll follow these steps:
1. Identify given data and constants:
- The enthalpy of vaporization ([tex]\(\Delta H_{\text{vap}}\)[/tex]) of water is 40.7 kJ/mol.
- The mass of water (H2O) is 27.9 g.
- The molar mass of water (H2O) is approximately 18.015 g/mol.
2. Calculate the number of moles of H2O:
Using the molar mass of H2O, we can convert the mass of water to moles. The formula to find the number of moles ([tex]\(n\)[/tex]) is:
[tex]\[
n = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
Substituting the given values:
[tex]\[
n = \frac{27.9 \text{ g}}{18.015 \text{ g/mol}} \approx 1.5487 \text{ mol}
\][/tex]
3. Calculate the quantity of heat released:
The quantity of heat ([tex]\(q\)[/tex]) released when water condenses can be found using the formula:
[tex]\[
q = n \, \Delta H_{\text{vap}}
\][/tex]
Substituting the values for [tex]\(n\)[/tex] and [tex]\(\Delta H_{\text{vap}}\)[/tex]:
[tex]\[
q = 1.5487 \text{ mol} \times 40.7 \text{ kJ/mol}
\][/tex]
[tex]\[
q \approx 63.03 \text{ kJ}
\][/tex]
Thus, the quantity of heat released when 27.9 g of H2O condenses is approximately [tex]\(63.03 \text{ kJ}\)[/tex].
Among the given options:
- [tex]\(60.00 \text{ kJ}\)[/tex]
- [tex]\(61.05 \text{ kJ}\)[/tex]
- [tex]\(63.09 \text{ kJ}\)[/tex]
- [tex]\(68.60 \text{ kJ}\)[/tex]
The closest option to our calculated value (63.03 kJ) is [tex]\(63.09 \text{ kJ}\)[/tex].
Therefore, the correct answer is [tex]\(63.09 \text{ kJ}\)[/tex].
