nCa(HCO3)2 =0.1975mol
nCa(OH)2 = 0.3 mol
PTHH
Ca(HCO3)2 + Ca(OH)2--> CaCO3 + H2O
0.1975-------------------------0.1975
=> m = 19.75g
=>nCO2 = 0.1975 mol =>VCO2 =4.424l
2HCl + Ca(OH)2-->CaCl2 + 2H2O
0.205----0.1025 mol
=> C% =\(\frac{0.1025\cdot111}{37.4125+7.585}\)=0.25M