Second order reaction with different reactants
The general solution, if the reaction is $aA+bB \to product$, is to allow an amount $x$ to react at time $t$ then there will be concentrations $(A_0-ax)$ and $(B_0-bx)$ of species A and B respectively at time $t$ if $A_0,B_0$ are the concentrations at time zero.  The rate equation is $\displaystyle  -\frac{1}{a}\frac{d(A_0-ax)}{dt}=\frac{dx}{dt}=k(A_0-ax)(B_0-bx)$ which can be integrated ($0\to x, 0\to t)$ to give

$$ \ln\left(\frac{A_0-ax}{B_0-bx} \right)=k(A_0b-B_0a)t +\ln\left(\frac{A_0}{B_0} \right)$$

If the reaction follows this scheme plot the left hand side vs $t$ and this should produce a straight line of slope $k(A_0b-B_0a)$.

Should it happen that concentrations are initially $A_0/B_0 = a/b$ so that species are present in stoichiometric amounts then as $A_0b-B_0a=0$ the  solution above cannot be used. In this case the rate equation becomes $\displaystyle  \frac{dA}{dt}=-bkA^2$ which gives $\displaystyle \frac{1}{A}-\frac{1}{A_0}=bkt$.