Ammonia, NH3, is used as a refrigerant. At its boiling point of -33 degress Celcius, the enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is required to vaporize 355 g of ammonia at -33 degrees Celcius?

a. 1.12 kJ
b. 152 kJ
c. 251 kJ
d. 486 kJ
e. 8.27 x 10^3 kJ

i thought is was D

thx

To calculate the heat required to vaporize a given amount of a substance, we can use the formula:

Q = n * ΔH

Where:
Q is the heat required (in joules)
n is the number of moles of the substance
ΔH is the enthalpy of vaporization (in joules per mole)

First, let's convert the mass of ammonia (355 g) into moles. To do this, we need to know the molar mass of ammonia, which is 17.03 g/mol.

Number of moles of ammonia = Mass / Molar mass
Number of moles of ammonia = 355 g / 17.03 g/mol

Next, we can substitute the values into the formula:

Q = n * ΔH
Q = (355 g / 17.03 g/mol) * 23.3 kJ/mol

Now, we need to convert the unit of kilojoules to joules, since the molar enthalpy is given in kilojoules and we want the heat in joules.

Q = (355 g / 17.03 g/mol) * (23.3 kJ/mol * 1000 J/kJ)

Finally, we can calculate the value:

Q = (355 / 17.03) * (23.3 * 1000)
Q ≈ 486000 J

Therefore, the heat required to vaporize 355 g of ammonia at -33 degrees Celsius is approximately 486,000 joules. Converting this to kilojoules, we have 486 kJ.

So the answer is (d) 486 kJ.

q = mass x heat vaporization = ??

Convert 355 g NH3 to mols and plug in for mass. heat of vaporization is in kJ/mol. The answer will be in kJ. Note the correct spelling of Celsius.