Khí thoát ra là CO2
\(nCO_2=\frac{0,448}{22,4}=0,02\left(mol\right)\)
PTHH
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
0,02..........0,04.........0,02......................
\(C\%ddHCl=\frac{0,04\cdot36,5}{400}\cdot100\%=0,365\%\)
mdd sau phản ứng = \(0,02\cdot100+400-0,02\cdot44=401,12\left(g\right)\)
\(C\%ddspu=\frac{0,02\cdot111}{401,12}\cdot100\%\approx0,55\%\)