Answered

Get the information you need quickly and easily with IDNLearn.com. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.

How much heat energy is required to boil 66.7 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol.

Sagot :

Dsdrajlinidn

Answer:

91.7 kJ

Explanation:

Step 1: Given data

  • Mass of ammonia (m): 66.7 g
  • Molar heat of vaporization of ammonia (ΔH°vap): 23.4 kJ/mol

Step 2: Calculate the moles (n) corresponding to 66.7 g of ammonia

The molar mass of ammonia is 17.03 g/mol.

66.7 g × 1 mol/17.03 g = 3.92 mol

Step 3: Calculate the heat (Q) required to boil 3.92 moles of ammonia

We will use the following expression.

Q = ΔH°vap × n

Q = 23.4 kJ/mol × 3.92 mol = 91.7 kJ

We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.