Question

When menthol, 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 combination of methanol has the balanced, thermochemical equation CH3OH(g) + 3/2O2(g) — CO2(g) + 2H2O(1) DeltaH= -764Kj How much methanol, in grams, must be found to produce 701 kj of heat?

Answer 1

We have the following balanced equation:

$CH_3OH_{(g)}+\frac{3}{2}O_{2\text{ (g)}}\rightarrow CO_{2\text{ (g)}}+2H_2O_{(l)}$

They also give us the heat of reaction equal to -764 kJ, i.e. it is an exothermic reaction.

By observing the reaction, we can deduce that for this heat to be generated, one mole of methanol is needed. Now let's see how many grams that mole of methanol equals. We will use the molecular weight equal to 32.04 g/mol

$\begin{gathered} g\text{ of methanol = 1 mol }\times\text{ 32.04 g/mol} \\ g\text{ of methanol }=\text{ 32.04 g} \end{gathered}$

Now we know the grams of methanol that generate 764 kJ, because the heat of reaction is directly proportional to the mass of the reactants, we can apply a rule of three to know the grams needed to produce a heat of reaction equal to 701 kJ:

$\begin{gathered} \frac{32.04\text{ g of methanol}}{764\text{ kJ}}=\text{ }\frac{x\text{ g of methanol}}{701\text{ kJ}} \\ x\text{ g of methanol = }\frac{32.04\text{ g of methanol }\times\text{ 701 kJ}}{764\text{ kJ}} \\ g\text{ of methanol = 29.4 g} \end{gathered}$

So, 29.4 g of methanol must be found to produce 701 kJ of heat