How to write the balanced chemical equation for the system in a rate plot?

I have the rate plot below:


[![rate plot](https://i.stack.imgur.com/2pWHS.jpg)](https://i.stack.imgur.com/2pWHS.jpg)


I know that $\ce{A}$ and $\ce{B}$ are the reactants, and $\ce{C}$ is the product. The textbook says that the balanced equation is $$\ce{3A + B -> 2C}\\[1em]$$ However, I am not sure how to determine those coefficients based on the graph. Any help is greatly appreciated.




By stoichiometry, the number of moles of species $j$ is related to the extent of the reaction via
\begin{align}
 n\_j &= n\_{j0} + \nu\_j \varepsilon \hspace{3 cm} (\text{divide by $V$}) \\
 \frac{n\_j}{V} &= \frac{n\_{j0}}{V\_0} + \nu\_j \frac{\varepsilon}{V} \\
 C\_j &= C\_{j0} + \nu\_j \varepsilon\_V
 \rightarrow \nu\_j = \frac{C\_j - C\_{j0}}{\varepsilon\_V} \tag{1} \\
\end{align}
where we assumed that the volume of the reacting mixture does not change as the reaction take place, i.e., $V\approx V\_0$. This is a good approximation for liquid-phase reactions, where the temperature does not change much, so that the density is almost constant. We also defined for convenience an extent of reaction per unit volume $\varepsilon := \varepsilon\_V / V$


Apply Eq. (1) to each component
\begin{align}
 \nu\_A = \frac{C\_A - C\_{A0}}{\varepsilon\_V} \tag{2} \\
 \nu\_B = \frac{C\_B - C\_{B0}}{\varepsilon\_V} \tag{3} \\
 \nu\_C = \frac{C\_C - C\_{C0}}{\varepsilon\_V} \tag{4} \\
\end{align}
Dividing Eq. (2) with Eq. (3), and Eq. (4) with Eq. (3), and using the values shown in your plot, we obtain the ratio of the stoichiometric numbers
\begin{align}
 \require{cancel}
 \frac{\nu\_A}{\nu\_B} & = \frac{C\_A - C\_{A0}}{C\_B - C\_{B0}} =
 \frac{(1.2 - 6) \; \cancel{\pu{M}}}{(2.4 - 4) \; \cancel{\pu{M}}} = 3
 \rightarrow \nu\_A = 3 \nu\_B \tag{5} \\
 \frac{\nu\_C}{\nu\_B} &= \frac{C\_C - C\_{C0}}{C\_B - C\_{B0}} =
 \frac{(3.2 - 0) \; \cancel{\pu{M}}}{(2.4 - 4) \; \cancel{\pu{M}}} = -2
 \rightarrow \nu\_C = -2\nu\_B \tag{6}
\end{align}
And finally we write the chemical equation using Eqs. (5) and (6)
\begin{align}
 \require{cancel}
 \nu\_A A + \nu\_B C + \nu\_C C = 0 \\
 3\cancel{\nu\_B} A + \cancel{\nu\_B} B - 2\cancel{\nu\_B} C = 0 \\
 3 A + B - 2 C = 0 \rightarrow \boxed{3A + B \rightarrow 2C}
\end{align}


 0