Question

2C3H7OH(l) + 9O2(g) → 8H2O(g) + 6CO2(g)

The combustion of isopropanol, i.e. rubbing alcohol, releases 1,830 kJ/mol.
Given that the molecular weight of isopropanol is 60.096 g/mol, calculate the enthalpy of combustion for 45.0g of isopropanol.

A)
147 kJ


B)
1,370 kJ


C)
2,160 kJ


D)
4,950 kJ

Answer 1

B) 1,370 kJ.

Step-by-step explanation:

• For the reaction:

2C₃H₇OH(l) + 9O₂(g) → 8H₂O(g) + 6CO₂(g) , ΔH = - 1,830 kJ/mol.

• Firstly, we need to calculate the no. of moles of 45.0 g of isopropanol:

n = mass/molar mass = (45.0 g)/(60.096 g/mol) = 0.7488 mol.

• Using cross multiplication:

1.0 mol of isopropanol releases → 1,830 kJ.

∴ 0.7488 mol of isopropanol releases → ??? kJ.

∴ The enthalpy of combustion for 45.0g of isopropanol = (0.7488 mol)(1830 kJ)/(1.0 mol) = 1370 kJ.

So, the right choice: B) 1,370 kJ.