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.