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

The answer is 3.23 mol.

Given data:

Mass in gram of MgCl2 = 308

Molecular mass = 95.21 g/mol

Number of moles = ?

Formula:

Number of moles = Mass in grams / Molecular mass

By putting the values in above formula:

Number of moles = 308 / 95.21

                           = 3.23 mol

So if the weight of MgCl2 is 308 grams. So by the formula of number of moles, it will contain 3.23 moles. Mass in grams of MgCl2 is divided by its molecular mass to get number of moles of MgCl2.