$a,PTPƯ:ZnO+2CH_3COOH\xrightarrow{} (CH_3COO)_2Zn+H_2O$
$b,n_{ZnO}=\dfrac{16,2}{81}=0,2mol.$
$Theo$ $pt:$ $n_{(CH_3COO)_2Zn}=n_{ZnO}=0,2mol.$
$⇒m_{(CH_3COO)_2Zn}=0,2.183=36,6g.$
$c,Theo$ $pt:$ $n_{CH_3COOH}=2n_{ZnO}=0,4mol.$
$⇒C\%_{CH_3COOH}=\dfrac{0,4.60}{200}.100\%=12\%$
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