7/

a,

$\text{Gọi số mol}:\\Etanol:C_{2}H_{5}OH:x\\Phenol:C_{6}H_{5}OH:y\\C_{2}H_{5}OH+Na \to C_{2}H_{5}ONa+1/2H_{2}\\C_{6}H_{5}OH+Na \to C_{6}H_{5}ONa+1/2H_{2}\\nH_{2}=\frac{2,24}{22,4}=0,1\\⇒1/2x+1/2y=0,1(1)\\C_{6}H_{5}OH+3HNO_{3} \to C_{6}H_{2}(NO_{2})_{3}OH+3H_{2}O\\nC_{6}H_{5}OH=nC_{6}H_{2}(NO_{2})_{3}OH=\frac{22,9}{229}=0,1\\⇒y=0,1(2) \\ \text{Thay (2) vào (1)}⇒x=0,1\\\%mC_{2}H_{5}OH=\frac{0,1.46}{0,1.46+0,1.94}.100=32,85\%\\\%mC_{6}H_{5}OH=100-32,85=67,15\%$

b,

$C_{6}H_{6}+Cl_{2} \buildrel{{t^o,Fe}}\over\longrightarrow C_{6}H_{5}Cl+HCl \\ C_{6}H_{5}Cl+NaOHđ \buildrel{{t^o\text{ cao , p cao}}}\over\longrightarrow C_{6}H_{5}ONa+NaCl+H_{2}O\\C_{6}H_{5}ONa+CO_{2}+H_{2}O \to C_{6}H_{5}OH+3HCl\\nC_{6}H_{5}OH=nC_{6}H_{6}=0,1\\nC_{6}H_{6}\text{ban đầu}=\frac{0,1}{78\%}=\frac{5}{39}\\mC_{6}H_{6}\text{ban đầu}=\frac{5}{39}.78=10g$

8/

a,

$Phenol:C_{6}H_{5}OH\\ \text{Ancol no đơn chức}: C_{n}H_{2n+1}OH \\C_{6}H_{5}OH+3Br_{2} \to C_{6}H_{2}Br_{3}OH+3HBr\\nC_{6}H_{5}OH=nC_{6}H_{2}Br_{3}OH=\frac{33,1}{331}=0,1\\\%mC_{6}H_{5}OH=\frac{0,1.94}{21,4}.100≈44\% \\\%mAncol=100-44=56\%$

b,

$C_{6}H_{5}OH+Na \to C_{6}H_{5}ONa+1/2H_{2} \\\text{ 0,1                                                   0,05} \\C_{n}H_{2n+1}OH+Na \to C_{n}H_{2n+1}ONa +1/2H_{2} \\\text{0,2                                                              0,1} \\ nH_{2}=\frac{3,36}{22,4}=0,15 \\ nH_{2}\text{phản ứng với ancol}=0,15-0,05=0,1\\ mAncol=21,4-0,194=12g \\ MAncol=\frac{12}{0,2}=60 \\ ⇒14n+2+16=60 \\ ⇔n=3 \\⇒CTPT:C_{3}H_{7}OH$