A thermite charge for welding a rail contains 2.70 kg of aluminium powder and the exact amount of…

Step 1. Moles of aluminium. $2700 / 26.98 = 100.1\ \text{mol}$.

Step 2. Iron(III) oxide. Half as many moles: $50.04\ \text{mol}$. The mass is $50.04 \times 159.69 = 7990\ \text{g} = 7.99\ \text{kg}$.

Step 3. Iron. One mole of iron for each mole of aluminium: $100.1 \times 55.85 = 5589\ \text{g} = 5.59\ \text{kg}$.

Step 4. The enthalpy of the equation. $\Delta H = -1675.7 - (-824.2) = -851.5\ \text{kJ}$ for $2$ mol of Al. Elements have zero enthalpy of formation.

Step 5. The heat. $50.04 \times 851.5 = 42\,600\ \text{kJ} = 42.6\ \text{MJ}$.

Source. The enthalpies of formation are standard values (NIST Chemistry WebBook, CRC Handbook).