CaCO3+HCl---->CaCl2+H2O+CO2
CO2+Ba(OH)2---->BaCO3+H2O
n BaCO3=31,51/171=0,16(mol)
n Ba(OH)2=0,5.0,4=0,2(mol)
----> Ba(OH)2 dư..chỉ xảy ra 1 muối BaCO3
Theo pthh2
n CO2=n BaCO3=0,16(mol)
Theo pthh1
n CaCO3=n CO2=0,16(mol)
m CaCO3=0,16.100=16(g)