nCO=nCO2=0,2=nCuO pứ
nCa(OH)2=0,05 mol
nHCl=0,4 mol
CuO+CO=>Cu+CO2
Cr sau pứ gồm Cu, CuO dư
CO2+Ca(OH)2=>CaCO3+H2O
CuO+2HCl=>CuCl2+H2O(2)
0,15<=0,3 mol
2HCl+Ca(OH)2=>CaCl2+2H2O
0,1<=0,05 mol
=>nHCl(2)=0,3 mol
=>nCuO du=0,15 mol
=>tổng CuO bđ=0,15+0,2=0,35 mol=>m=0,35.80=28gam