When sodium bromate(V), ${\text{NaBr}}{{\text{O}}_{\text{3}}}$, is heated, it reacts according to the equation below.

$${\text{2NaBr}}{{\text{O}}_{\text{3}}}{\text{(s)}} \to {\text{2NaBr(s)}} + {\text{3}}{{\text{O}}_{\text{2}}}{\text{(g)}}$$

What amount, in mol, of ${\text{NaBr}}{{\text{O}}_{\text{3}}}$ produces ${\text{2.4 d}}{{\text{m}}^{\text{3}}}$ of oxygen gas, measured at room temperature and pressure? (Molar volume of gas $= {\text{24 d}}{{\text{m}}^{\text{3}}}{\text{mo}}{{\text{l}}^{ - 1}}$ at room temperature and pressure.)

A.     0.017

B.     0.067

C.     0.10

D.     0.15

B