High School

How many grams of ethanol, C2H5OH, can be boiled with 187.2 kJ of heat energy? The molar heat of vaporization of ethanol is 38.6 kJ/mol.

Answer :

The grams of ethanol, C₂H₅OH, can be boiled with 187.2 kJ of the heat energy. The molar heat of vaporization of ethanol is 38.6 kJ/mol is 222.9 g.

The heat energy of the ethanol, C₂H₅OH = 187.2 kJ

The molar heat of vaporization of ethanol, C₂H₅OH = 38.6 kJ /mol

The expression is as follows :

Q = n H

Where,

H = heat of vaporization of ethanol = 38.6 kJ /mol

Q = heat energy of the ethanol, C₂H₅OH = 187.2 kJ

n = moles

n = 187.2 / 38.6

n = 4.84 mol

The mass of the ethanol = moles × molar mass

The mass of the ethanol = 4.84 mol × 46.07 g/mol

The mass of the ethanol = 222.9 g

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Using the molar heat of vaporization for ethanol, which is 38.6 kJ/mol, it can be calculated that 223.10 grams of ethanol can be boiled with 187.2 kJ of heat energy.

The question asks how many grams of ethanol, C₂H₅OH, can be boiled with 187.2 kJ of heat energy, given the molar heat of vaporization of ethanol is 38.6 kJ/mol.

To solve this, we use the equation:

  1. Calculate the number of moles of ethanol that can be vaporized by the provided energy.
  2. Multiply the number of moles by the molar mass of ethanol to find the mass in grams.

Step 1: Number of moles = 187.2 kJ × 38.6 kJ/mol = 4.85 mol

Step 2: Mass in grams = 4.85 mol × 46 g/mol = 223.10 g

Therefore, 223.10 grams of ethanol can be boiled with 187.2 kJ of heat energy.