$1$ C6H6 + Cl2 -> C6H5Cl + HCl

$2$ C6H5Cl + NaOH$H2Oр-р$ -> C6H5OH + NaCl

Стехиометрическая схема: C6H6 -> C6H5OH

n$C6H6$ = m$C6H6$/M$C6H6$ = 390/78 = 5$моль$

по стехиометрической схеме: nтеор$C6H5OH$ = n$C6H6$ = 5$моль$

mтеор$C6H5OH$ = nтеор$C6H5OH$*M$C6H5OH$ = 5*94 = 470$г$

ωвых$C6H5OH$ = $mпр(C6H5OH)/mтеор(C6H5OH)$*100% = $410/470$*100% ≈ 87,2%

Ответ: ωвых$C6H5OH$ = 87,2%