\(\text{nCaCO3=}\frac{10}{\text{40+12+16.2}}\text{=0,1(mol)}\)
\(\text{PTHH: CO2 + Ca(OH)2}\rightarrow\text{ CaCO3 + H2O}\)
Tỉ lệ:____0,1_________________________0,1___(mol)
\(\text{V CO2 = 0,1.22,4=2,24(l)}\)
Ca(OH)2+CO2--->CaCO3+H2O
CaCO3+CO2+H2O--->Ca(HCO2)2
Ta có
n caCO3=10/100=0,1(mol)
n Ca(OH)2=0,3.0,5=0,15(mol)
Do n Ca(OH)2>n CaCO3 --->xảy ra 1 muối trung hòa
Theo pthh1
n CO2=n CaCO3=0,19mol)
V CO2=0,1.22,4=2,24(l)