$a/PTHH :$
$2CH_3COOH+CaCO_3\to (CH_3COO)_2Ca+CO_2+H_2O$
$b/n_{CO_2}=0,448/22,4=0,02mol$
$\text{Theo pt :}$
$n_{CH_3COOH}=2.n_{CO_2}=2.0,02=0,04mol$
$⇒m_{CH_3COOH}=0,04.60=2,4g$
$C\%CH_3COOH=\dfrac{2,4.100\%}{40}=6\%$
$\text{c/Theo pt :}$
$n_{CaCO_3}=n_{CO_2}=0,02mol$
$⇒m_{CaCO_3}=0,02.100=2g$