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.

one negative thing it that it can kill water animals. another negative thing is it can cause the destruction of ecosystem. one way water pollution can be reduced is to conserve energy.