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$a)$

`n_(Mg` `=` `m/M` `=` `(2,4)/(24)=0,1(mol)`

PTHH: $Mg$ $+$ $2HCl$ `->` $MgCl_2$ $+$ $H_2$
Theo PT: `n_(HCl` `=` `2n_(Mg` `=` `0,2(mol)`

`m_(HCl` `=` `n.M=0,2.36,5=7,3(g)`

`m_(ddHCl` `=` `(m_{HCl}.100)/(C%)=(7,3.100)/(14,6)=50(g)`

`V_(ddHCl` `=` `m/D` `=` `(50)/(1,1)≈45,45(ml)`

$b)$

Theo PT: `n_(MgCl_2` `=` `n_(Mg` `=` `0,1(mol)`

Đổi `45,45(ml)``=0,04545(l)`

`C_MMgCl_2` `=` `n/V` `=` `(0,1)/(0,04545)≈2,2(M)`

`m_(MgCl_2` `=` `n.M=0,1.95=9,5(g)`

Theo PT: `n_(H_2` `=` `n_(Mg` `=` `0,1(mol)`

`m_(ddMgCl_2` `=` `2,4+50-0,1.2=52,2(g)`

`C%_(MgCl_2` `=` `(m_{MgCl_2})/(m_{ddMgCl_2}).100%=(9,5)/(52,2).100%≈18,2%`

Đáp án + Giải thích các bước giải:

 a)

`Mg + 2HCl->MgCl_2 + H_2↑`

`n_{Mg}=m/M=(2,4)/(24)= 0,1 (mol)`

Theo PT: `n_{HCl}=2n_{Mg}=2 xx 0,1= 0,2 (mol)`

`=> m_{\text{dd HCl}}=(0,2 xx 36,5)/(14,6\text{%}) xx 100\text{%}= 50 (g)`

`=> V_{\text{dd HCl}}=m/D=(50)/(1,1)= 45,45 (ml) = 0,04545 (ml)`

b)

Theo PT: `n_{H_2}=n_{Mg}= 0,1 (mol)`

`=> m_{\text{dd sau pứ}}=2,4 + 50 - 0,1 xx 2= 52,2 (g)`

Theo PT: `n_{MgCl_2}=n_{Mg}= 0,1 (mol)`

`=> C_{M_{MgCl_2}}=n/V=(0,1)/(0,04545)= 2,2 (M)`

`=> \text{C%}_{MgCl_2}=(0,1 xx 95)/(52,2) xx 100\text{%}= 18,2\text{%}`