$n_{CaCO_3} = \dfrac{20}{100} = 0,2(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Theo PTHH :
$n_{HCl} = 0,4(mol) ; n_{CO_2} = 0,2(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,4.36,5}{14,6\%} = 100(gam)$
$m_{dd\ sau\ pư} = 20 + 100 - 0,2.44 = 111,2(gam)$
$C\%_{CaCl_2} = \dfrac{0,2.111}{111,2}.100\% = 19,96\%$