CO2+ Ca(OH)2 --> CaCO3 + H2O
Ta có: CM=\(\frac{n}{V}\)=> nCa(OH)2= 1.0,05 =0,05 mol
n= \(\frac{V}{22,4}\)=> nCO2= \(\frac{1,344}{22,4}=0,06\) mol
Vì nCO2 > nCa(OH)2 nên CO2 dư
Từ pt; nCaCO3=nCa(OH)2=0,05 mol
n=\(\frac{m}{M}\)=> mCaCO3=0,05.100= 5 mol