Câu trả lời:
nCaCO3=20100=0,2<nCO2=0,4nCaCO3=20100=0,2<nCO2=0,4
Do đó, sản phầm có tạo muối axit
Ca(OH)2+CO2→CaCO3+H2OCa(OH)2+CO2→CaCO3+H2O
Ca(OH)2+2CO2→Ca(HCO3)2Ca(OH)2+2CO2→Ca(HCO3)2
nCO2=2nCa(HCO3)2+nCaCO3nCO2=2nCa(HCO3)2+nCaCO3
⇒nCa(HCO3)2=0,4−0,22=0,1(mol)⇒nCa(HCO3)2=0,4−0,22=0,1(mol)
nCa(OH)2=nCaCO3+nCa(HCO3)2=0,3(mol)nCa(OH)2=nCaCO3+nCa(HCO3)2=0,3(mol)
mdd Ca(OH)2=D.V=1,05.400=420(gam)mdd Ca(OH)2=D.V=1,05.400=420(gam)
C%Ca(OH)2=0,3.74420.100%=5,29%