Đáp án:
\(H\% = 69,33\% \)
Giải thích các bước giải:
$\text{PTHH:}$ \({C_6}{H_6} + C{l_2}\xrightarrow{{Fe,{t^o}}}{C_6}{H_5}Cl + HCl\)
$\text{Ta có:}\ {n_{{C_6}{H_5}Cl}} = \dfrac{{78}}{{112,5}} \approx 0,6933\ mol$
$\text{Theo PTHH:}\ {n_{{C_6}{H_6}\ pư}} = {n_{{C_6}{H_5}Cl}} \to {m_{{C_6}{H_6}\ pư}} = 54,08\ g$
\(\to H\% = \dfrac{{{n_{{C_6}{H_6}pư}}}}{{{n_{{C_6}{H_6}\ bđ}}}}.100\% = \dfrac{{54,08}}{{78}}.100\% = 69,33\% \)