Ammonia, NH3, is used as a refrigerant. At its boiling point of -33ºC, the enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is required to vaporize 125 g of ammonia at -33ºC?
171kj
q=molesx heat of vaporization
The above formula is correct but you will need to change 125 g NH3 to moles to use it.
23.3X125(1 mole NH3/17.024g(molar mass of NH3)= 171kj
To calculate the heat required to vaporize a given mass of ammonia, we can follow these steps:
Step 1: Convert the given mass of ammonia from grams to moles.
To do this, we need to know the molar mass of ammonia (NH3), which is calculated as follows:
Molecular mass of Nitrogen (N) = 14.01 g/mol
Molecular mass of Hydrogen (H) = 1.01 g/mol
Total molecular mass of ammonia (NH3) = (1 × 14.01 g/mol) + (3 × 1.01 g/mol) = 17.03 g/mol
Now we can calculate the number of moles:
Number of moles = Mass (in grams) / Molar mass (in grams per mole)
Given mass = 125 g
Number of moles = 125 g / 17.03 g/mol
Step 2: Calculate the heat required to vaporize the moles of ammonia.
Now that we have the number of moles of ammonia, we can calculate the heat required to vaporize it using the enthalpy of vaporization.
Enthalpy of vaporization = 23.3 kJ/mol
Heat required = Enthalpy of vaporization × Number of moles
Step 3: Convert the heat into the desired unit, kJ.
The heat is given in Joules (J), so we need to convert it to kilojoules (kJ).
1 kJ = 1000 J
Heat required (in kJ) = Heat required (in J) / 1000
By following these steps, you will obtain the amount of heat required to vaporize 125 g of ammonia at -33ºC.