In a reaction $2\mathrm{A}+\mathrm{B}⟶{\mathrm{A}}_2\mathrm{B}$, the reactant ‘A’ will disappear at

According to reaction $2\mathrm{A}+\mathrm{B}⟶{\mathrm{A}}_2\mathrm{B}$,

The rate of the reaction is

$\begin{array}{rcl}-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{A}\right]& =& -\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{B}\right]=+\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{A}}_2\mathrm{B}\right]\\ \therefore -\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{A}\right]& =& -2\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{B}\right]\end{array}$

Therefore the rate at which A disappears will be twice the rate at which B disappears.