A thermochemical equation from Enthalpy and Thermochemical Equations is also a conversion factor. "$\Delta H = -890.3\ \text{kJ}$ per mole of reaction" lets you ask how much heat a real amount of fuel gives, using the same moves as Mass-to-Mass Stoichiometry.
Key idea
$$q = (\text{moles of reaction}) \times \Delta H, \qquad \text{moles of reaction} = \frac{\text{moles of a substance}}{\text{its coefficient}}.$$ So: grams → moles of the substance → divide by its coefficient → multiply by $\Delta H$. A negative result is heat released.
Why it works
"Per mole of reaction" means per one run of the equation exactly as written. If the equation is $2\,\mathrm{H_2} + \mathrm{O_2} \to 2\,\mathrm{H_2O}$ with $\Delta H = -571.6\ \text{kJ}$, one run uses 2 mol of $\mathrm{H_2}$. Burning 1 mol of $\mathrm{H_2}$ is half a run, so it releases half of 571.6 kJ. The coefficient is the bridge between the substance you have and the reaction's $\Delta H$, just as it was the bridge between two substances in stoichiometry.
Worked example: Heat from burning methane
$\mathrm{CH_4} + 2\,\mathrm{O_2} \to \mathrm{CO_2} + 2\,\mathrm{H_2O(l)}$, $\Delta H = -890.3\ \text{kJ}$. How much heat is released by burning $10.0\ \text{g}$ of $\mathrm{CH_4}$ ($16.05\ \text{g/mol}$)?
- Moles: $10.0/16.05 = 0.6231\ \text{mol}\ \mathrm{CH_4}$, which is $0.6231$ mol of reaction (coefficient 1).
- $q = 0.6231 \times (-890.3) = -555\ \text{kJ}$.
Answer: $555\ \text{kJ}$ released.
Worked example: A coefficient of 2
$2\,\mathrm{H_2} + \mathrm{O_2} \to 2\,\mathrm{H_2O(l)}$, $\Delta H = -571.6\ \text{kJ}$. How much heat does burning $1.00\ \text{g}$ of $\mathrm{H_2}$ ($2.02\ \text{g/mol}$) release?
- Moles of $\mathrm{H_2}$: $1.00/2.02 = 0.4950\ \text{mol}$.
- Moles of reaction: $0.4950/2 = 0.2475$.
- $q = 0.2475 \times (-571.6) = -141\ \text{kJ}$.
Answer: $141\ \text{kJ}$ released, the highest of any fuel per gram.
Worked example: Grams needed
$\mathrm{C_3H_8} + 5\,\mathrm{O_2} \to 3\,\mathrm{CO_2} + 4\,\mathrm{H_2O(l)}$, $\Delta H = -2219.9\ \text{kJ}$. What mass of propane ($44.11\ \text{g/mol}$) must burn to release $1000\ \text{kJ}$?
- Moles of reaction: $1000/2219.9 = 0.4505$; propane's coefficient is 1, so $0.4505\ \text{mol}$.
- Mass: $0.4505 \times 44.11 = 19.9\ \text{g}$.
Answer: $19.9\ \text{g}$
Dividing by molar mass turns these into fuel values per gram: methane $890.3/16.05 = 55.5\ \text{kJ/g}$, propane $50.3\ \text{kJ/g}$, hydrogen $141.5\ \text{kJ/g}$. Per gram, hydrogen is the richest fuel, though a gram of it takes up a great deal of space as a gas.
Common mistake
Multiplying $\Delta H$ by grams instead of moles, or by the moles of a substance whose coefficient isn't 1. In the hydrogen example, $1.00 \times 571.6$ or $0.4950 \times 571.6$ would both be wrong. Check: always convert to moles of reaction, dividing by the coefficient, before you multiply by $\Delta H$.