$n_{Cl_2}=\dfrac{22,4}{22,4}=1(mol)$
$n_{NaOH}=0,1.2=0,2(mol)$
$n_{Ca(OH)_2}=0,1.0,5=0,05(mol)$
$2NaOH+Cl_2\to NaCl+NaClO+H_2O$
$2Ca(OH)_2+2Cl_2\to CaCl_2+Ca(ClO)_2+2H_2O$
Ta có: $\dfrac{n_{NaOH}}{2}+n_{Ca(OH)_2}<n_{Cl_2}$
Nên $Cl_2$ dư.
$n_{NaCl}=\dfrac{n_{NaOH}}{2}=0,1(mol)$
$n_{CaCl_2}=\dfrac{n_{Ca(OH)_2}}{2}=0,025(mol)$
$\to m_{\text{muối clorua}}=0,1.58,5+0,025.111=8,625g$