Ammonia,NH₃,is used as a refrigerant.At its boiling point of -33 $\circ$ C,the standard enthalpy of vaporization of ammonia is 23.3 kJ/mol.How much heat is released when 50.0 g of ammonia is condensed at -33 $\circ$ C?

Question

Quizplus · Accepted Answer

First, we need to calculate how many moles of ammonia are in 50.0g.
 
50.0g NH3 x (1mol NH3/17.03g NH3) = 
2.936 mol NH3

Then, we use the formula:

q = n*ΔH

where q is the heat released, n is the number of moles, and ΔH is the enthalpy of vaporization (-23.3 kJ/mol, but we change the sign to indicate heat being released).

q = 2.936 mol NH3 * (-23.3 kJ/mol) = 
-68.4 kJ

Therefore, the answer is D, -68.4 kJ.

Ammonia,NH3,is Used as a Refrigerant ∘\circ∘ C,the Standard Enthalpy of Vaporization of Ammonia Is 23

Ammonia,NH₃,is Used as a Refrigerant $\circ$ C,the Standard Enthalpy of Vaporization of Ammonia Is 23