Đề là: `C_2 H_5 OH` và `CH_3 COOH` nhỉ?
-Giải-
`a)PTHH:`
`C_2 H_5 OH+K->C_2 H_5 OK+1/2H_2 \uparrow`
`CH_3 COOH+K->CH_3 COOK+1/2H_2 \uparrow`
`b)n_[H_2]=[5,6]/[22,4]=0,25(mol)``
Gọi `n_[C_2 H_5 OH]=x;n_[CH_3 COOH]=y`
`=>` $\begin{cases} 46x+60y=25,8\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,25 \end{cases}$
`<=>` $\begin{cases} x=0,3\\y=0,2 \end{cases}$
`@m_[C_2 H_5 OH]=0,3.46=13,8(g)`
`@m_[CH_3 COOH]=25,8-13,8=12(g)`