Ta có: nH2 = 0,25 (mol)
\(CH_3OH+K\rightarrow CH_3OK+\dfrac{1}{2}H_2\)
\(C_6H_5OH+K\rightarrow C_6H_5OK+\dfrac{1}{2}H_2\)
⇒ nCH3OH + nC6H5OH = 2nH2 = 0,5 (1)
\(n_{KOH}=\dfrac{200.8,4\%}{56}=0,3\left(mol\right)=n_{C_6H_5OH}\) (2)
\(C_6H_5OH+KOH\rightarrow C_6H_5OK+H_2O\)
Từ (1) và (2) ⇒ nCH3OH = 0,2 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3OH}=\dfrac{0,2.32}{0,2.32+0,3.94}.100\%\approx18,5\%\\\%m_{C_6H_5OH}\approx81,5\%\end{matrix}\right.\)