`n_(CO_2)=5.6/22.4=0.25(mol)`
`PTHH:CO_2+Ba(OH)_2->BaCO_3↓+H_2O`
`Theo pt:n_(Ba(OH)_2)=n_(BaCO_3)=n_(CO_2)=0.25(mol)`
`a, C_M(ddBa(OH)_2)=0.25/0.1=2.5M`
`b, m_(BaCO_3)=0.25·197=49.25g`
`c,` Trung hòa `ddBa(OH)_2` bằng `ddHCl`
`Ba(OH)_2+2HCl->BaCl_2+2H_2O`
`Theo pt:n_(HCl)=2n_(Ba(OH)_2)=0.25·2=0.5(mol)`
`m_(ddHCl)=(0.5·36.5·100%)/(20%)=91.25g`