How much methanol, in grams, must be burned to produce 683 kJ of heat?

Question

When methanol, CH3OH, is burned in the presence of oxygen gas, 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

How much methanol, in grams, must be burned to produce 683 kJ of heat?

mass: ?g

Answer 1

Hey there! So first off, looking at the equation you listed, methanol produces 764 kJ of heat per mole.

Given the stoichiometric coefficient in the balanced equation (there is an implied "1" in the balanced equation next to methanol, one mole of methanol produces 764 kJ of heat.

So, we'll need to work backwards and find out how many moles methanol produces 683 kJ of heat. If you like ratio and proportion, you could set it up like this...

1 mole/764 kJ = x moles/683 kJ.

Solving for x, we see that 0.894 moles of methanol produces 683 kJ of heat.

Now, let's answer the question which states we must give our answer in grams of methanol. Since methanol's formula is CH3OH that's (1 C = 12.01, + 1 O = 16, + 4 H = (4 x 1.01) = 32.05 g/mol.

One last conversion! Let's convert 0.894 moles to grams...

0.894 moles x 32.05 g/1 mol = 28.6527 grams.

Since the numbers given all contained 3 sig figs, let's now trim our answer to 3 sig figs...

28.7 grams of methanol.

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Mike