Đáp án:

Giải thích các bước giải:

$n_{NaOH} = 0,1.2 = 0,2(mol)$

$n_{Ca(OH)_2} = 0,1.0,5 = 0,05(mol)$

$n_{Cl_2} = 1(mol)$

$Cl_2 + 2NaOH \to NaCl + NaClO + H_2O$

$n_{NaCl} = n_{Cl_2\ pư} =  \dfrac{1}{2}n_{NaOH} = 0,1(mol)$

$2Cl_2 + 2Ca(OH)_2 \to Ca(ClO)_2 + CaCl_2 + 2H_2O$
$n_{CaCl_2} =  \dfrac{1}{2}n_{Ca(OH)_2} = 0,025(mol)$

Vậy muối clorua gồm :$NaCl:0,1(mol)\\ CaCl_2:0,025(mol)$

$\Rightarrow m_{muối} = 0,1.58,5 + 0,025.111 = 8,625(gam)$

$n_{Cl_2}=\dfrac{22,4}{22,4}=1(mol)$

$n_{NaOH}=0,1.2=0,2(mol)$

$n_{Ca(OH)_2}=0,1.0,5=0,05(mol)$

$2NaOH+Cl_2\to NaCl+NaClO+H_2O$

$2Ca(OH)_2+2Cl_2\to CaCl_2+Ca(ClO)_2+2H_2O$

Ta có: $\dfrac{n_{NaOH}}{2}+n_{Ca(OH)_2}<n_{Cl_2}$

Nên $Cl_2$ dư.

$n_{NaCl}=\dfrac{n_{NaOH}}{2}=0,1(mol)$

$n_{CaCl_2}=\dfrac{n_{Ca(OH)_2}}{2}=0,025(mol)$

$\to m_{\text{muối clorua}}=0,1.58,5+0,025.111=8,625g$