`MgCl_2+2NaOH→Mg(OH)_2+2NaCl`

`n_(MgCl_2)=(190.20)/(100.95)=0,4(mol)`

`n_(NaOH)=(200.10)/(100.40)=0,5(mol)`

`n_(NaOH):2<n_(MgCl_2)`

`→MgCl_2` `dư`

`a)`

`n_(Mg(OH)_2)=(0,5)/2=0,25(mol)`

`m_(Mg(OH)_2)=0,25.58=14,5(g)`

`b)`

`n_(MgCl_2) dư=0,4-0,25=0,15(mol)`

`mdd sau pứ=190+200-14,5=375,5(g)`

`C%_(MgCl_2) dư=(0,15.95.100)/(375,5)=3,79%`

`C%_(NaCl)=(0,5.58,5.100)/(375,5)=7,79%`

\(n_{MgCl_2^{}}^{} = \frac{{190.20\% }}{{95}} = 0,4{\text{ mol}}\)

\(n_{NaOH}^{} = \frac{{200.10\% }}{{40}} = 0,5\;{\text{mol}}\)

\(MgCl_2^{} + 2NaOH \to 2NaCl + Mg(OH)_2^{}\)

\({\text{   }}\frac{{0,4}}{1} > \frac{{0,5}}{2} \to MgCl_2^{}{\text{ du}}{\text{, tinh theo NaOH}}\)

\({\text{   0}}{\text{,25       0}}{\text{,5           0}}{\text{,5            0}}{\text{,25}}\)

\({\text{m}}_{Mg(OH)_2^{}}^{} = 0,25.58 = 14,5g\)

Đề thiếu V

\(C_{M_{MgCl_2^{}}^{}}^{} = \frac{{0,4 - 0,25}}{V}\)

\(C_{M_{NaCl}^{}}^{} = \frac{{0,5}}{V}\)