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 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

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.