\(n_{CaCO_3}=\dfrac{20}{100}=0.2\left(mol\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(0.2............0.4.............0.2............0.2\)
\(C\%_{HCl}=\dfrac{0.4\cdot36.5}{208.8}\cdot100\%=7\%\)
\(m_{dd}=20+208.8-0.2\cdot44=220\left(g\right)\)
\(C\%_{CaCl_2}=\dfrac{0.2\cdot111}{220}\cdot100\%=10.09\%\)