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`a)`

`200ml=0,2l`

`->n_{KOH}=0,2.1=0,2(mol)`

`CH_3COOH+KOH->CH_3COOK+H_2O`

Theo PT: `n_{CH_3COOH}=n_{KOH}=0,2(mol)`

`->\%m_{CH_3COOH}={0,2.60}/{16,6}.100\%≈72,29\%`

`->\%m_{C_2H_5OH}=100-72,29=27,71\%`

`b)

`n_{C_2H_5OH}={16,6-0,2.60}/{46}=0,1(mol)`

`C_2H_5OH+Na->C_2H_5ONa+1/2H_2`

`CH_3COOH+Na->CH_3COONa+1/2H_2`

Theo PT: `n_{H_2}=1/2(n_{C_2H_5OH}+n_{CH_3COOH})=0,15(mol)`

`->V_{H_2}=0,15.22,4=3,36(l)`

`c)`

`n_{CH_3COOC_2H_5}={4,4}/{88}=0,05(mol)`

Do `0,1<0,2->H` tính theo `C_2H_5OH`

`C_2H_5OH+CH_3COOH`  $\buildrel{{t^o,H^+}}\over\rightleftharpoons$  `CH_3COOC_2H_5+H_2O`

Theo PT: `n_{C_2H_5OH(t t)}=n_{CH_3COOC_2H_5}=0,05(mol)`

`->H={0,05}/{0,1}.100\%=50\%`