Đáp án:

$\%n_{C_6H_5OH}=8,33\%$

Giải thích các bước giải:

Thí nghiệm 1: 

$C_6H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\ (1)\\CH_3OH+Na\to CH_3ONa+\dfrac{1}{2}H_2\ (2)\\n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

Thí nghiệm 2:

$C_6H_5OH+3Br_2\to C_6H_2Br_3OH+3HBr\\n_{C_6H_2Br_3OH}=\dfrac{16,55}{331}=0,05(mol)\\n_{C_6H_5OH}=n_{C_6H_2Br_3OH}=0,05(mol)\\\to n_{H_2\ (1)}=\dfrac{1}{2}n_{C_6H_5OH}=0,025(mol)\\\to n_{H_2\ (2)}=0,3-0,025=0,275(mol)\\\to n_{CH_3OH}=2n_{H_2\ (2)}=0,55(mol)\\n_X=n_{C_6H_5OH}+n_{CH_3OH}=0,05+0,55=0,6(mol)\\\%n_{C_6H_5OH}=\dfrac{0,05}{0,6}\times 100\%=8,33\%$

\(C_6^{}H_5^{}OH + 3Br_2^{} \to C_6^{}H_2^{}Br_3^{}OH + 3HBr\)

\(n_{C_6^{}H_2^{}Br_3^{}OH}^{} = \frac{{16,55}}{{331}} = 0,05\;{\text{mol}}\)

\( \to {\text{n}}_{C_6^{}H_5^{}OH}^{} = 0,05\;{\text{mol}}\)

\(C_6^{}H_5^{}OH + Na \to C_6^{}H_5^{}ONa + \frac{1}{2}H_2^{}\)

\(\;{\text{   0}}{\text{,05                                      0}}{\text{,025}}\)

\(CH_3^{}OH + Na \to CH_3^{}ONa + \frac{1}{2}H_2^{}\)

\(\;{\text{ x                                         }}\frac{1}{2}x\)

\(n_{H_2^{}}^{} = 0,3 = \frac{1}{2}x + 0,025\)

\( \to x = 0,55\;{\text{mol}}\)

\(\% n_{C_6^{}H_5^{}OH}^{} = \frac{{0,05}}{{0,05 + 0,55}}.100 = 8,33\% \)