$MgO~+~2HCl-->MgCl_2~+~H_2O$

$MgCO_3~+~2HCl-->MgCl_2~+~H_2O~+~CO_2$

$\left \{ {{n_{MgO}~.~40~~+~~n_{MgCO_3}~.~84=12,4~(~g~)} \atop {n_{MgO}~+~n_{MgCO_3}=\frac{19}{24+71}=0,2~(~mol~)}} \right.$

        $\left \{ {{n_{MgO}=0,1~(~mol~)} \atop {n_{MgCO_3}=0,1~(~mol~)}} \right.$ 

`C%_{MgO}=\frac{0,1. 40. 100%}{12,4}≈32,26%`

`C%_{MgCO_3}=100%- 32,26%=67,74%`

.

$n_{HCl}=0,1~.~2~~+~~0,1~.~2=0,4~(~mol~)$

$m_{HCl}=n~.~M=0,4~.~36,5=14,6~(~g~)$

`m_{dd HCl}=m_{HCl}:C%=14,6:7,3%=200(g)`

$m_{dung~dịch~sau~p/ứ}=m_{dung~dịch~HCl}~+~m_{hh~bột}-m_{CO_2}$

                                                $=200~+~12,4~-~0,1~.~44=208~(~g~)$

`C%_{MgCl_2}=`$\frac{0,2~.~95~.~100}{208}≈9,13$`%`

`a)`

`MgO+2HCl->MgCl_2+H_2O`

`MgCO_3+2HCl->MgCl_2+CO_2+H_2O`

`Hh:MgO:x(mol);MgCO_3:y(mol)`

`40x+84y=12,4(1)`

`n_{MgCl_2}=x+y={19}/{95}=0,2(2)`

`(1)(2)->x=y=0,1`

`\%m_{MgO}={0,1.40}/{12,4}.100\%=32,26\%`

`\%m_{MgCO_3}=100-32,26=67,74\%`

`b)`

`n_{HCl}=2n_{MgCl_2}=0,4(mol);n_{CO_2}=y=0,1(mol)`

`m_{dd\ sau}=12,4+{0,4.36,5}/{7,3\%}-0,1.44=208(g)`

`C\%_{MgCl_2}={19}/{208}.100\%=9,13\%`