Question

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.

Answer 1

Answer: 1,702 kJ

Explanation: the math for the other one is correct just don't divide 1,830 by two, the number is already correct

Answer 2

850.95 kj

Step-by-step explanation:

Given data:

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