Đáp án:

`C%_{dd\ MgCl_2}~~12,34%`

Giải thích các bước giải:

`a)`

`Pt:MgO+2HCl->MgCl_2+H_2O`

`b)`

`n_{MgO}={8}/40=0,2(mol)`

`n_{HCl}={109,5.10%}/{36,5.100%}=0,3(mol)`

`Ta\ co: {0,2}/1>{0,3}/2=>HCl\ het\ , MgO\ du `

`Theo\ pt: n_{MgO(pu)}=n_{MgCl_2}=1/{2}n_{HCl}=0,15(mol)`

`n_{MgO(du)}=0,2-0,15=0,05(mol)`

`m_{dd\ spu}=8+109,5-(0,05.40)=115,5(g)`

`C%_{dd\ MgCl_2}={0,15.95}/{115,5}.100%~~12,34%`

`#Ly`

$a)$

PTHH: $MgO$ $+$ $2HCl$ `->` $MgCl_2$ $+$ $H_2O$

$b)$

`n_(MgO` `=` `(8)/(40)=0,2` $(mol)$

`m_(HCl` `=` `(109,5.10)/(100)=10,95` $(g)$

`n_(HCl` `=` `(10,95)/(36,5)=0,3` $(mol)$

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

Xét tỉ lệ: `(0,2)/(1)` $>$ `(0,3)/(2)`

`=>` $MgO$ dư, tính theo `n_(HCl` 

Theo PT, `n_(MgCl_2` `=` `(1)/(2)n_(HCl` `=` `0,15` $(mol)$

`m_(MgCl_2` `=` `0,15.95=14,25` $(g)$

Theo PT, `n_(MgO(pư)` `=` `(1)/(2)n_(HCl` `=` `0,15` $(mol)$

`n_(MgO(dư)` `=` `0,2-0,15=0,05` $(mol)$

`m_(MgO(dư)` `=` `0,05.40=2` $(g)$

`m_(dd(spư)` `=` `8+109,5-2=115,5` $(g)$

`C%_(MgCl_2` `=` `(14,25)/(115,5).100%≈12,34%`