Methanol ($\ce{CH3OH}$) is regarded by many chemists as a possible liquid fuel for the future. The combustion of methanol occurs according to the equation:

$$\ce{2CH3OH(g) + 3O2(g) -> 2CO2(g) + 4H2O(l)}\quad\Delta H=-1506\ \mathrm{kJ\ mol^{-1}}$$

(a) Determine the heat of combustion of methanol, in $\mathrm{kJ\ g^{-1}}$.

(b) If $0.750\ \mathrm{mol}$ of methanol is mixed with $0.750\ \mathrm{mol}$ of oxygen and the mixture is ignited, how much energy will be released?

I have some working for this question, but not sure if it is right, because there is two moles of methanol in the equation.

for 18a)

$$1506\ \mathrm{kJ}\space\text{for}\space1\ \mathrm{mol}$$ $$1=\frac{m}{M}$$ $$1=\frac{m}{32.04}$$ $$m=32.04\ \mathrm g$$ $$\therefore1506\ \mathrm{kJ}\space\text{for}\space32.04\ \mathrm g$$ $$\therefore47\ \mathrm{kJ}\space\text{for}\space1\ \mathrm g$$ $$\therefore\space\text{heat of combustion}=47\ \mathrm{kJ\ g^{-1}}$$

for 18b)

must use $0.75\ \mathrm{mol}$ of oxygen

$$\therefore0.75\times\frac{2}{3}=0.5\ \mathrm{mol}\space\text{of methanol}$$ $$1506\ \mathrm{kJ}\space\text{for}\space1\ \mathrm{mol}$$ $$1506\times0.5\space\text{for}\space0.5\ \mathrm{mol}$$ $$753\ \mathrm{kJ}\space\text{for}\space0.5\ \mathrm{mol}$$ Therefore, $753\ \mathrm{kJ}$ is released for $0.75\ \mathrm{mol}$ of oxygen and $0.75\ \mathrm{mol}$ of methanol. Only $0.5\ \mathrm{mol}$ of methanol can be used.

Not entirely sure if any of this working is correct, any help is appreciated.

  • $\begingroup$ Part (a) is correct. In part (b), you have correctly identified the limiting reactant as oxygen, and hence only 0.5 moles of methanol will be combusted. Since 2 moles of methanol yields 1506 kJ, 0.5 moles would yield 376.5 kJ or 377 kJ (to 3 sig.figs.) $\endgroup$ – Dr. J. Jun 17 '18 at 11:20
  • $\begingroup$ @Dr.J. Part (a) is not correct. The given enthalpy is for 2 mol of methanol. The molar enthalpy of combustion of liquid (!) methanol at 25 °C actually is 726.1 kJ/mol. $\endgroup$ – Loong Jun 17 '18 at 16:46
  • $\begingroup$ Correct answer for (a) should be 23.5 kJ/g as pointed out by Loong. Initial comment was made in haste. $\endgroup$ – Dr. J. Jun 18 '18 at 11:54

Part A

First, the molar mass (MM) of methanol must be calculated. Sum the atomic masses given on the periodic table when multiplied by the instances of each atom in methanol.

$$ \begin{split} \text{MM}_{\ce{CH3OH}} & = 1 \times 12.01~\mathrm{g} + 4 \times 1.01~\mathrm{g} + 1 \times 16.00~\mathrm{g} \\ & = 32.05~\mathrm{g} \end{split} $$

Employ dimensional analysis to change the units of the given quantity.

$$ \frac{-1506~\mathrm{kJ}}{1~\mathrm{mol_{rxn}}} \times \frac{1~\mathrm{mol_{rxn}}}{2~\mathrm{mol}_{\ce{CH3OH}}} \times \frac{1~\mathrm{mol}_{\ce{CH3OH}}}{32.05~\mathrm{g}} = -23.49~\frac{\mathrm{kJ}}{\mathrm{g}_{\ce{CH3OH}}} $$

The heat of combustion of methanol in $\mathrm{kJ}~\mathrm{g^{-1}}$ is -23.49.

Part B

Identify this as a limiting reagent problem. When given equimolar quantities of both reactants, the limiting reagent is the reactant with the greater coefficient in the balanced chemical equation ($\ce{O2 (g)}$ in this instance). Mathematically, determine the number of moles of $\ce{O2 (g)}$ needed to fully combust $0.750~\mathrm{mol}~\ce{CH3OH}$. If this number exceeds the provided moles of $\ce{O2 (g)}$, then $\ce{O2 (g)}$ is the limiting reagent.

$$ \frac{0.750~\mathrm{mol}_{\ce{CH3OH}}}{1} \times \frac{3~\mathrm{mol}_{\ce{O2}}}{2~\mathrm{mol}_{\ce{CH3OH}}} = 1.13~\mathrm{mol}_{\ce{O2}} $$

$$ 1.13~\mathrm{mol}_{\ce{O2}} > 0.750~\mathrm{mol}_{\ce{O2}} $$

Therefore, $\ce{O2 (g)}$ is the limiting reagent.

Knowing $\ce{O2 (g)}$ is the limiting reagent, employ dimensional analysis to find the quantity of energy released.

$$ \frac{0.750~\mathrm{mol}_{\ce{O2}}}{1} \times \frac{1~\mathrm{mol_{rxn}}}{3~\mathrm{mol}_{\ce{O2}}} \times \frac{1506~\mathrm{kJ}}{1~\mathrm{mol_{rxn}}} = 377~\mathrm{kJ} $$

377 kJ of energy is released by the given reaction in the listed scenario.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.