$$5$$ $$g$$ of benzene on nitration gave $$6.6$$ $$g$$ of nitrobenzene. The theoretical yield of the nitrobenzene will be

Solution :
$$\mathop {{C_6}{H_6}}\limits_{78\,g} + HN{O_3} \to \mathop {{C_6}{H_5}N{O_2}}\limits_{123\,g} + {H_2}O$$
Now since $$78g$$  of benzene on nitration give $$= 123g$$  nitrobenzene
hence $$5g$$  of benzene on nitration give
$$\frac{{123}}{{78}} \times 5 = 7.88g$$
nearest answer is (C) i.e. theoritical yield $$ = 7.88\,g$$