Identify Bronsted - Lowry acids in the reaction given below?

 $\underset{\mathrm{A}}{[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}}+\underset{\mathrm{B}}{{\mathrm{HCO}}_3^{-}}\iff \underset{\mathrm{C}}{{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\right(\mathrm{OH}\left)\right]}^{2+}}+\underset{\mathrm{D}}{{\mathrm{H}}_2{\mathrm{CO}}_3}$

The correct answer is

In a reaction, bronsted lowry acid is the reactant that loses a proton and is taken up by bronsted base. Reactant A $[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$ loses one proton from one ${\mathrm{H}}_2\mathrm{O}$ in the reaction described above, creating product C $[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{OH}{]}^{2+}$.

Therefore, the correct option is (A).