Enthalpy of Formation problems using Hess's Law
Find the $\Delta H^\circ_f$ of the following reaction:
$$\ce{C + O2 -&gt; CO2}$$

We have our list of reactions with known enthalpies:
$\ce{SrO + CO2 -&gt; SrCO3} \quad \Delta H^\circ_f = \pu{−234 kJ}$
$\ce{2SrO -&gt; 2Sr + O2} \quad \Delta H^\circ_f = \pu{1184 kJ} $
$\ce{2SrCO3 -&gt; 2Sr + 2C + 3O2} \quad \Delta H^\circ_f = \pu{2440 kJ} $
Hess's Law takes advantage of the fact that enthalpy is a state function, looking at our reactions, we need to find a way to derive what we want:
Carbon dioxide is on the reactants side, however, we want it on the products, so we can run that reaction reverse; however we need to change the sign of enthalpy if we do that:
$\ce{SrCO3 -&gt; SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
We need to get rid of $\ce{SrO}$, thus, we use our second equation:
$\ce{2SrO -&gt; 2Sr + O2 } \quad \Delta H^\circ_f = \pu{1184 kJ} $
Since, there's only one strontium oxide in the first reaction, we divide by the integer 2 across the chemical equation including the enthalpy of formation:
$\ce{SrO -&gt; Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
Now, with equation 3, we need carbon and oxygen on the reactants side, and need to remove $\ce{SrCO3}$ and also divide again by 2, since we only have one strontium.
$\ce{Sr + C + 3/2O2 -&gt; SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $
Now we add up all our reactions and enthalpies, and cancel out common elements on both sides:
$\ce{SrCO3 -&gt; SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
$\ce{SrO -&gt; Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
$\ce{Sr + C + 3/2O2 -&gt; SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $

$\ce{C + O2 -&gt; CO2} \quad \Delta H^\circ_f= (\pu{234 kJ}) + (\pu{592 kJ}) + (\pu{-1220 kJ}) = \pu{-394 kJ}$