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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 :

dsdrajlin

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

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