$\text{n$_{H2}$=2,24:22,4=0,1(mol)}$

$\text{n$_{kết tủa}$=n$_{AgCl}$=57,4:143,5=0,4(mol)}$

$\text{Gọi x là số mol MgO }$

$\text{a,PTHH:}$

$\text{Mg+2HCl→MgCl$_{2}$+H$_{2}$}$  (1)

$\text{0,1  ←0,2    ← 0,1  ←0,1 (mol)}$

$\text{MgO+2HCl→MgCl$_{2}$+H$_{2}$O}$  (2)

$\text{x          2x         x  (mol)}$

$\text{MgCl$_{2}$+2AgNO$_{3}$→2AgCl↓+Mg(NO3)$_{2}$}$  (3)

$\text{0,2   ←                  0,4  (mol)  }$

$\text{b,Từ PTHH (1) ta suy ra : n$_{Mg}$=0,1(mol)}$

$\text{⇒m$_{Mg}$=0,1.24=2,4(g)}$

$\text{Từ PTHH (3) ta suy ra : n$_{MgCl2}$=0,2(mol)}$

$\text{Mặt khác từ PTHH (1) và (2) ta suy ra :}$

$\text{n$_{MgCl2}$=0,1+x}$

$\text{⇒0,1+x=0,2}$

$\text{⇒x=0,1(mol)}$

$\text{⇒m$_{MgO}$=0,1.40=4(g)}$

$\text{⇒m=m$_{Mg}$+m$_{MgO}$=2,4+4=6,4(g)}$

$\text{%m$_{Mg}$=$\frac{2,4}{6,4}$.100%=37,5% }$

$\text{⇒%m$_{MgO}$=(100-37,5)%=62,5%}$

$\text{c,Từ PTHH (1) và (2) suy ra : n$_{HCl}$=0,2+2x=0,2+2.0,1=0,4(mol)}$

$\text{⇒m$_{HCl}$=0,4.36,5=14,6(g)}$

$\text{⇒m$_{dd}$=14,6:10%=146(g)}$

$\text{%m$_{MgCl2}$=$\frac{0,2.95}{146+6,4-0,1.2}$.100% ≈12,48%}$

$a)PTHH:Mg+2HCl\to MgCl_2+H_2(1)$

$MgO+2HCl\to MgCl_2+H_2O(2)$

$MgCl_2+2AgNO_3\to Mg(NO_3)_2+2AgCl\downarrow(3)$

$b)n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$n_{AgCl}=\dfrac{57,4}{143,5}=0,4(mol)$

Theo PT(3): $n_{MgCl_2}=0,5.n_{AgCl}=0,2(mol)$

Theo PT(1): $n_{Mg}=n_{H_2}=0,1(mol)\Rightarrow m_{Mg}=0,1.24=2,4(g)$

Theo PT(1) và PT(2): $n_{Mg}+n_{MgO}=n_{MgCl_2}$

$\Rightarrow 0,1+n_{MgO}=0,2$

$\Rightarrow n_{MgO}=0,1(mol)$

$\Rightarrow m_{MgO}=0,1.40=4(g)$

$\Rightarrow \%m_{Mg}=\dfrac{2,4}{2,4+4}.100\%=37,5\%$

$\Rightarrow \%m_{MgO}=100\%-37,5\%=62,5\%$

$c)$ Theo PT(1) và PT(2): $\Sigma n_{HCl}=2n_{MgCl_2}=0,4(mol)$

$\Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{10\%}=146(g)$

$m_{H_2}=0,1.2=0,2(g);m_{MgCl_2}=0,2.95=19(g);m_{hh}=4+2,4=6,4(g)$

$\Rightarrow C\%_{MgCl_2}=\dfrac{19}{6,4+146-0,2}.100\%\approx 12,48\%$