$n_{CaCO_3}=10/100=0,1mol$
$m_{HCl}=200.2\%=4g$
$⇒n_{HCl}=4/36,5=0,109mol$
$PTHH :$
$CaCO_3 + 2HCl\to H_2O+CO_2+CaCl_2↑$
$\text{Theo pt : 1 mol 2 mol}$
$\text{Theo đbài: 0,1 mol 0,109mol}$
$\text{⇒Sau pư CaCO3 dư}$
$\text{Theo pt :}$
$n_{CO_2}=1/2.n_{HCl}=1/2.0,109=0,0545mol$
$⇒m_{CO_2}=0,0545.44=2,398g$