Câu 26: C

$2C_2H_5OH+2Na\xrightarrow\ 2C_2H_5ONa+H_2\\ 2C_6H_5OH+2Na\xrightarrow\ 2C_6H_5ONa+H_2\\ n_{C_2H_5OH}:a(mol)\ ; n_{C_6H_5OH}:b(mol)\\ n_{H_2}=\dfrac{3,7185}{24,79}=0,15(mol)\\ \sum n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+\dfrac{1}{2}n_{C_6H_5OH}\\ \xrightarrow\ \dfrac{1}{2}a+\dfrac{1}{2}b=0,15\ (1)\\ C_6H_5OH+3Br_2\xrightarrow\ C_6H_2Br_3OH+3HBr\\ n_{C_6H_5OH}=n_{C_6H_2Br_3OH}=\dfrac{33,1}{331}=0,1(mol)\\ \xrightarrow\ b=0,1\ (2)\\ (1)(2)\xrightarrow\ a=0,2(mol)\\ \%m_{C_2H_5OH}=\dfrac{0,2.46}{0,2.46+0,1.94}.100\%=49,46\%$

Câu 27:

$\xrightarrow{btnt\ H:} n_H=2n_{H_2O}=\dfrac{2.3,24}{18}=0,36(mol)\\ n_O=2n_{H_2}=\dfrac{2.1,4874}{24,79}=0,12(mol)\\ n_{C(ancol)}=n_O=0,12(mol)\\m=m_C+m_H+m_O\\=0,12.12+0,36.1+0,12.16=3,72(g)$

Đáp án + Giải thích các bước giải:

Câu 26: C$\\$ Giả sử $n_{C_2H_5OH}=a(mol)\ ; n_{C_6H_5OH}=b(mol)$$\\$ $2C_2H_5OH+2Na\xrightarrow\ 2C_2H_5ONa+H_2$$\\$ $2C_6H_5OH+2Na\xrightarrow\ 2C_6H_5ONa+H_2$$\\$ $n_{H_2}=\dfrac{3,7185}{24,79}=0,15(mol)$$\\$ $ n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+\dfrac{1}{2}n_{C_6H_5OH}$$\\$ $\to \dfrac{1}{2}a+\dfrac{1}{2}b=0,15\ (1)$$\\$ $C_6H_5OH+3Br_2\xrightarrow\ C_6H_2Br_3OH+3HBr$$\\$ $n_{C_6H_2Br_3OH}=\dfrac{33,1}{331}=0,1(mol)$$\\$ $n_{C_6H_5OH}=n_{C_6H_2Br_3OH}=0,1(mol)$$\\$ $\to b=0,1(mol)\ (2)$$\\$ $(1)(2)\xrightarrow\ a=0,2(mol)$$\\$ $\%m_{C_2H_5OH}=\dfrac{0,2.46}{0,2.46+0,1.94}.100\%=49,46\%$$\\$ Câu 27:$\\$ Bảo toàn H:$\\$ $n_H=2n_{H_2O}=\dfrac{2.3,24}{18}=0,36(mol)$$\\$ $n_{H_2}=\dfrac{1,4874}{24,79}=0,06(mol)$$\\$ Mà $n_{C}=n_O=2n_{H_2}=0,12(mol)$$\\$ $m=m_C+m_H+m_O\\=0,12.12+0,36.1+0,12.16=3,72(g)$$\\$ $\to$ Không có đáp án phù hợp