$a,PTPƯ:Zn+2CH_3COOH\xrightarrow{} (CH_3COO)_2Zn+H_2↑$
$n_{Zn}=\dfrac{13}{65}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$b,Theo$ $pt:$ $n_{CH_3COOH}=2n_{Zn}=0,4mol.$
$⇒m_{CH_3COOH}=0,4.60=24g.$
$⇒m_{ddCH_3COOH}=\dfrac{24}{10\%}=240g.$
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