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

B)  1,370 kJ.

Explanation:

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

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

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.

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