Question

Methanol (CH₃OH) burns according to the equation:

2CH₃OH(l) + 3O₂(g) ----> 2CO₂(g) + 4H₂O(l), DH°ᵣₓₙ = –1454 kJ.
HINT: Note that DH°rxn = –1454 kJ heat is released when 2 moles of methanol is burned.
How much heat, in kilojoules (kJ), is given off when 85.0 g of methanol is burned ?

Answer 1

The heat released when 85.0 g of methanol is burned is approximately -1929.83 kJ.

Step-by-step explanation:

To calculate the heat of combustion of methanol (CH₃OH), we need to use the equation: 2CH₃OH(l) + 3O₂(g) ----> 2CO₂(g) + 4H₂O(l), with a DH°ᵣₓₙ value of -1454 kJ. We are given that 2 moles of methanol release -1454 kJ of heat. So, for 1 mole of methanol, the heat released would be -1454 kJ / 2 = -727 kJ. To find the heat released when 85.0 g of methanol is burned, we need to convert grams to moles. The molar mass of methanol is 32.04 g/mol, so 85.0 g of methanol is equivalent to 85.0 g / 32.04 g/mol = 2.653 moles of methanol. Therefore, the heat released when 85.0 g of methanol is burned would be -727 kJ/mol x 2.653 mol = -1929.83 kJ (rounded to three significant figures).