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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:
- Calculate the number of moles of ethanol that can be vaporized by the provided energy.
- 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.
