$a,PTPƯ:Na_2CO_3+2CH_3COOH\xrightarrow{} 2CH_3COONa+H_2O+CO_2↑$
$n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol.$
$Theo$ $pt:$ $n_{CH_3COOH}=2n_{CO_2}=0,4mol.$
$⇒m_{ddCH_3COOH}=\dfrac{0,4.60}{6\%}=400g.$
$b,Theo$ $pt:$ $n_{Na_2CO_3}=n_{CO_2}=0,2mol.$
$⇒m_{Na_2CO_3}=0,2.106=21,2g.$
$c,Theo$ $pt:$ $n_{CH_3COONa}=2n_{CO_2}=0,4mol.$
$⇒C\%_{ddCH_3COONa}=\dfrac{0,4.82}{21,2+400-(0,2.44)}.100\%=7,95\%$
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