Question

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , 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)+3/2O2(g) ⟶ CO2(g) + 2H2O(I) Δ H = − 764 kJ.

How much methanol, in grams, must be burned to produce 575 kJ of heat?

Answer 1

24.1 g

Step-by-step explanation:

Let's consider the following thermochemical equation.

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

In order to produce 764 kJ of heat, 1 mole of methanol must be burned. To produce 575 kJ, the required moles of methanol are:

-575 kJ × (1 mol CH₃OH/ -764 kJ) = 0.753 mol CH₃OH

The molar mass of methanol is 32.04 g/mol. 0.753 moles of methanol represent a mass of:

mass = moles × molar mass

0.753 mol × (32.04 g/mol) = 24.1 g