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

ΔHreaction = 1,830 kJ/mol

Calculate the enthalpy of combustion for 55.9g isopropanol given the molecular weight of isopropanol is 60.096 g/mol.

850.95 kj

Explanation:

Given dа ta:

ΔH of reaction = 1830 kj/mol

Enthalpy of combustion for 55.9 g of isopropanol = ?

Solution:

Number of moles of isopropanol:

Number of moles = mass / molar mass

Number of moles = 55.9 g / 60.096 g/mol

Number of moles = 0.93 mol

1830 kj heat is produced when 2 mole of isopropanol react.

For one mole:

1830 kj/2 = 915 kj/mol

For 0.93 mol

0.93 mol * 915 kj/mol = 850.95 kj