Question

When methanol, CH3OH,CH3OH, is burned in the presence of oxygen gas, O2,O2, a large amount of heat energy is released. For this reason, it is often used as a fuel in high performance racing cars. The combustion of methanol has the balanced, thermochemical equation CH3OH(g)+32O2(g)⟶CO2(g)+2H2O(l)ΔH=−764 kJ CH3OH(g)+32O2(g)⟶CO2(g)+2H2O(l)ΔH=−764 kJ How much methanol, in grams, must be burned to produce 665 kJ665 kJ of heat?

Answer 1

27.87 g of methanol

Step-by-step explanation:

The correct equation for the reaction is;

CH3OH(g) + 3/2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH=−764 kJ

We must put down the thermo chemical equation for this reaction clearly because it will guide our work in the solution of the problem.

We can see from the equation that 1 mole of methanol produces 764KJ of heat.

x moles of methanol will produce 665KJ of heat

x= 665 KJ×1 mole / 764KJ

x= 0.87 moles of methanol

Molar mass of methanol= 32.04 g/mol

Mass of methanol required= number of moles of methanol × molar mass of methanol

Mass of methanol= 0.87moles × 32.04 g/mol

Mass of methanol= 27.87 g of methanol