Question

Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of n-propanol (C3H7OH)is burned in air, -33.4 kJ of heat is liberated. Calculate the heat of combustion of n-propanol in kJ/mol.

Answer 1

Answer : The heat of combustion of n-propanol is 0.554 kJ/mol

Explanation :

First we have to calculate the moles of n-propanol.

$\text{Moles of n-propanol}=\frac{\text{Mass of n-propanol}}{\text{Molar mass of n-propanol}}$

Molar mass of n-propanol = 60.09 g/mole

$\text{Moles of n-propanol}=\frac{1.00g}{60.09g/mole}=0.0166mole$

Now we have to calculate the heat of combustion of n-propanol.

As, 0.0166 mole of n-propanol liberated heat of combustion = -33.4 kJ

So, 1 mole of n-propanol liberated heat of combustion = 0.0166 × (-33.4 kJ)

                                                                                           = 0.554 kJ/mol

Therefore, the heat of combustion of n-propanol is 0.554 kJ/mol