`Mg+2HCl→MgCl_2+H_2`

`a)`

`n_(Mg)=(9,6)/24=0,4(mol)`

`V_(H_2)=0,4.22,4=8,96(l)`

`b)`

`m_(HCl)=0,4.2.36,5=29,2(g)`

`mdd_(HCl)=(29,2.100)/(7,3)=400(g)`

`c)`

`m_(MgCl_2)=0,4.95=38(g)`

`mdd spứ=9,6+400-0,4.2=

`C%MgCl_2=(38.100%)/(408,8)=9,295%`

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{9,6}}{{24}} = 0,4\,mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,4\,mol\\
{V_{{H_2}}} = 0,4 \times 22,4 = 8,96l\\
b)\\
{n_{HCl}} = 2{n_{Mg}} = 0,8\,mol\\
{m_{HCl}} = 0,8 \times 36,5 = 29,2g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{29,2}}{{7,3\% }} = 400g\\
c)\\
{m_{{\rm{dd}}spu}} = 9,6 + 400 - 0,4 \times 2 = 408,8g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,4\,mol\\
{C_\% }MgC{l_2} = \dfrac{{0,4 \times 95}}{{408,8}} \times 100\%  = 9,3\% 
\end{array}\)