Now since \(78\,g\) of benzene on nitration give \(= 123\,g\) nitrobenzene
hence \(5\,g\) of benzene on nitration give
\( = \frac{{123}}{{78}} \times 5 = 7.88\,g\)
nearest answer is \((c)\) i.e. theoritical yield
\(=7.88\,g\)
$I.\,SeO_3^{2 - } + BrO_3^ - + {H^ + } \to SeO_4^{2 - } + B{r_2} + {H_2}O$
$II.\,BrO_3^ - + AsO_2^ - + {H_2}O \to B{r^ - } + AsO_4^{3 - } + {H^ + }$