chemical equilibrium reaction for burned gases in engine
It's simpler than it might look. As a random example:

$$\ce{3A + 2B-&gt;2C + 4D}$$

If I were to ask you how much of $\ce{D}$ would be produced after $5\  mol$ of $\ce{B}$ were consumed, then you would immediately answer $10\  mol$ "due to the stoichiometry". In more elaborate terms, for a $\Delta n_B=n_B^{after}-n_B^{before}=-5\ mol$, one has $\Delta n_D=n_D^{after}-n_D^{before}=+10\ mol$. Any infinitesimal change in quantity of any species $i$ ($\delta n_i$) is accompanied by a shift in the quantities of all other species precisely determined by the stoichiometry of the reaction (the stoichiometric coefficients, $v_i$), such that $\frac{\delta n_i}{v_i}$ is constant for all species $i$. For the above reaction, $-\frac{\delta n_A}{3}=-\frac{\delta n_B}{2}=\frac{\delta n_C}{2}=\frac{\delta n_D}{4}$. The ratio $\frac{\delta n_i}{v_i}$ is defined as $\delta n$, and you can think of it as a counter for "how many times the reaction happened" ($\delta n&gt;0$ if left to right, $\delta n&lt;0$ if right to left).

$\delta n$ is a general change in amount of substances, but the particular case where changes are calculated relative to the equilibrium condition ($\Delta n=n^{equilibrium}-n^{before}$), then it represents the extent of reaction.