= Hess's law equation cancellation
{c}
Keep the <stoichiometry> attached to the <enthalpy>. Use the same abstract $A,B,C$ and suppose each of two entries fixes an <amount of substance> for every species:
$$
\begin{aligned}
R_1:&\quad 2A\longrightarrow2B, &&\Delta H_1=+36\,\mathrm{kJ},\\
R_2:&\quad C\longrightarrow B, &&\Delta H_2=+25\,\mathrm{kJ}.
\end{aligned}
$$
Here each unit coefficient represents one <mole> on the common basis. To obtain $A\to C$, halve $R_1$ and reverse $R_2$. Adding those transformed <chemical equations> gives
$$
\begin{aligned}
\tfrac12R_1:&\quad A\longrightarrow B, &&\tfrac12\Delta H_1=+18\,\mathrm{kJ},\\
-R_2:&\quad B\longrightarrow C, &&-\Delta H_2=-25\,\mathrm{kJ},\\
\text{sum}:&\quad A+B\longrightarrow B+C, &&\Delta H=-7\,\mathrm{kJ}.
\end{aligned}
$$
The equal amounts of $B$ on opposite sides cancel, leaving $A\to C$. Doubling this final <chemical equation> would give $2A\to2C$ and $\Delta H=-14\,\mathrm{kJ}$; its change per <mole> of $A$ would remain $-7\,\mathrm{kJ\,mol^{-1}}$.
More generally, write each <chemical equation> as a <stoichiometric vector> $\boldsymbol\nu_j$, with negative entries for <reactants> and positive entries for <reaction products>. Finding coefficients $a_j$ with $\boldsymbol\nu_{\rm target}=\sum_j a_j\boldsymbol\nu_j$ also gives $\Delta H_{\rm target}=\sum_j a_j\Delta H_j$, using compatible state and amount conventions. An identical <chemical species> cancels only with the same <thermodynamic phase> and composition conditions. A leftover <water> or <dilution> step is a real part of the balance.
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