$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$
$CaO+2HCl\to CaCl_2+H_2O$
$n_{CO_2}=\dfrac{1,12}{22,4}=0,05(mol)$
$\Rightarrow n_{CaCO_3}=0,05(mol)$
$m_{CaCO_3}=0,05.100=5g$
$\Rightarrow m_{CaO}=10-5=5g$
$n_{CaO}=\dfrac{5}{56}(mol)$
$\Rightarrow n_{HCl}=2n_{CaCO_3}+2n_{CaO}=\dfrac{39}{140}(mol)$
$m_{dd HCl}=\dfrac{39}{140}.36,5:20\%=50,84g$
$\Rightarrow V_{HCl}=\dfrac{50,84}{1,2}=42,367ml$
$n_{CaCl_2}=\dfrac{n_{HCl}}{2}=\dfrac{39}{280}(mol)$
$\Rightarrow m_{CaCl_2}=111.\dfrac{39}{280}=15,46g$