Molar heat of combustion with different fuels I'm not sure exactly how to attack this question, I'm sure that you have to use equations such as 

$$Q=mc\,\Delta T$$
$$n=\frac{M}{m}$$

But I don't know where to use them. Any help would be appreciated

I just attempted the question, I'm not sure if this is the correct working.

Assuming 1 mol of each substance

Ethyne: 
$$\begin{align}n&amp;=\frac{M}{m}\\[6pt]
1&amp;=\frac{26.038\ \mathrm{g/mol}}{m}\\[6pt]
m&amp;=26.038\ \mathrm g\end{align}$$
producing 1630 kJ

Hexane:
Same working as above so:
$$m=86.178\ \mathrm g$$
producing 1300 kJ

Ethane:
$$m=30.07\ \mathrm g$$
producing 1560 kJ

Simplifying each of the mass and energy relationships for each substance gives:

Ethyne: 1 gram for 62.6 kJ

Hexane: 1 gram for 15.09 kJ

Ethane: 1 gram for 51.89 kJ

This would imply that ethyne would be the best fuel, as it uses the least amount of mass and the greatest amount of energy is produced.

Not sure if this is correct. Any tips would be great.
This question is asking which substance has the greatest heat density, or energy released per mass combusted (something like J/g).

$\ce{C2H2}$ has a mass of $26.04\ \mathrm{g/mol}$. Since the $\Delta H$ in its combustion is $1\,630\ \mathrm{kJ/mol}$, you will be looking at a heat density of $\frac{1\,630\ \mathrm{kJ/mol}}{26.04\ \mathrm{g/mol}} = 62.6\ \mathrm{\frac{kJ}{g}}.$

Similarly, for $\ce{C6H14}$, you get $\frac{1\,300\ \mathrm{kJ/mol}}{86.178\ \mathrm{g/mol}} = 15.1\ \mathrm{\frac{kJ}{g}}.$

For $\ce{C2H6}$, you get $\frac{1\,560\ \mathrm{kJ/mol}}{30.07\ \mathrm{g/mol}} = 51.9\ \mathrm{\frac{kJ}{g}}.$

It looks like you were on the right track.