`a,`

`n_(MgCl_2)=0,2.0,15=0,03(mol)`

`MgCl_2+2NaOH->Mg(OH)_2+2NaCl`

$Mg(OH)_2\xrightarrow{t^o}MgO+H_2O$

Theo PT

`n_(NaOH)=2n_(MgCl_2)=0,06(mol)`

`n_(MgO)=n_(MgCl_2)=0,03(mol)`

`=>m_(MgO)=0,03.40=1,2(g)`

`b,`

Theo PT

`n_(NaCl)=2n_(MgCl_2)=0,06(mol)`

`=>C_(MNaCl)=\frac{0,06}{0,5}=0,12M`

Đáp án:

\(\begin{array}{l} a,\ m=m_{MgO}=1,2\ g.\\ b,\ C_{M_{NaCl}}=0,12\ M.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:\\ MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl\ (1)\\ Mg(OH)_2\xrightarrow{t^o} MgO+H_2O\ (2)\\ n_{MgCl_2}=0,2\times 0,15=0,03\ mol.\\ Theo\ pt\ (1):\ n_{Mg(OH)_2}=n_{MgCl_2}=0,03\ mol.\\ Theo\ pt\ (2):\ n_{MgO}=n_{Mg(OH)_2}=0,03\ mol.\\ \Rightarrow m=m_{MgO}=0,03\times 40=1,2\ g.\\ b,\\ V_{\text{dd spư}}=V_{\text{dd MgCl$_2$}}+V_{\text{dd NaOH}}=200+300=500\ ml=0,5\ lít.\\ Theo\ pt\ (1):\ n_{NaCl}=2n_{MgCl_2}=0,06\ mol.\\ \Rightarrow C_{M_{NaCl}}=\dfrac{0,06}{0,5}=0,12\ M.\end{array}\)

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