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?

Question

Carley Randolph

Chemistry

21 February, 03:13

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?

+1

Answers (1)

Know the Answer?

Nina Crane · Answer 1

24.1 g

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