n(CO2)=8.8/44=0.2mol
设CaCO3有xmol
CaCO3+2HCl===CaCl2+H2O+CO2↑
1/x=1/0.2
x=0.2
m(CaCO3)=0.2×100=20g
CaCO3%=20/25×100%=80%