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`MgO + 2HCl -> MgCl_2 + H_2 O`

`0,15`                      `0,15`                    `(mol)`

`CuO + 2HCl -> CuCl_2 + H_2 O`

 `0,2`                       `0,2`                       `(mol)`

`n_[HCl]=0,7.1=0,7(mol)`

Gọi `n_[MgO]=x;n_[CuO]=y`

`=>` $\begin{cases} 40x+80y=22\\2x+2y=0,7 \end{cases}$

`<=>` $\begin{cases} x=0,15\\=0,2 \end{cases}$

`a)%n_[MgO]=[0,15]/[0,15+0,2].100=42,86%`

`=>%n_[CuO]=100-42,86=57,14%`

`b)C_[M_[MgCl_2]]=[0,15]/[0,7]=0,2(M)`

   `C_[M_[CuCl_2]]=[0,2]/[0,7]=0,3(M)`

\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

  0,2          1,2             0,4                 ( mol )

\(m_{FeCl_3}=0,4.162,5=65\left(g\right)\)

\(m_{dd\left(HCl\right)}=\dfrac{1,2.36,5}{20\%}=219\left(g\right)\)

\(\%O=100\%-20,72\%-28,82\%=50,46\%\)

Giả sử: A có \(100\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=20,72\left(g\right)\\m_S=28,82\left(g\right)\\m_O=50,46\left(g\right)\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}n_{Na}=0,9\left(mol\right)\\n_S=0,9\left(mol\right)\\n_O=3,2\left(mol\right)\end{matrix}\right.\)

\(n_{Na}:n_S:n_O=0,9:0,9:3,2=2:2:7\)

`=>` CTHH A: \(Na_2S_2O_7\)

Đặt CTTQ: \(Na_xS_yO_z\)

\(x:y:z=\dfrac{20.72}{23}:\dfrac{28.82}{32}:\dfrac{\left(100-20.72-28.82\right)}{16}=0,90:0,90:3,15=1:1:3,5\left(không.thoả.mãn\right)\)

Vậy không tồn tại CTPT thoả mãn đề.

\(2M\left(OH\right)_n\rightarrow\left(t^o\right)M_2O_n+nH_2O\)

\(2MOH\rightarrow M_2O+H_2O\)

\(\Rightarrow n=1\)

\(n_{H_2SO_4}=\dfrac{100.9,8\%}{98}=0,1\left(mol\right)\)

`CuO+H_2SO_4->CuSO_4+H_2O`

  0,1                          0,1                      ( mol )

`m_{CuO}=0,1.80=8(g)`

`m_{CuSO_4}=0,1.160=16(g)`

CuO+H₂SO₄->CuSO₄+H₂O

0,1----0,1-------0,1-------0,1

m_H2SO4=9,8g

=>n_H2SO4=\(\dfrac{9,8}{98}\)0,1 mol

=>m _CuO=0,1.80=8g

=>m _CuSO4=0,1.160=16g

\(n_{BaCl_2}=\dfrac{200.10,4\%}{208}=0,1\left(mol\right)\) ; \(n_{AgNO_3}=\dfrac{400.68\%}{170}=1,6\left(mol\right)\)

\(BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\downarrow\)

 0,1            < 1,6                                         ( mol )

 0,1               0,2                0,1                 0,2    ( mol )

\(m_{ddspứ}=200+400-0,2.143,5=571,3\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{Ba\left(NO_3\right)_2}=\dfrac{0,1.261}{571,3}.100=4,56\%\\C\%_{AgNO_3\left(dư\right)}=\dfrac{\left(1,6-0,2\right).170}{571,3}.100=41,65\%\end{matrix}\right.\)

Quy hỗn hợp thành \(\left\{{}\begin{matrix}FeO\\Fe_2O_3\end{matrix}\right.\)

Ta có: \(\dfrac{n_{FeO}}{n_{Fe_2O_3}}=\dfrac{1}{1}\) \(\Rightarrow n_{FeO}=n_{Fe_2O_3}=x\left(mol\right)\)

\(\Rightarrow m_{hh}=72x+160x=23,2\)

\(\Leftrightarrow x=0,1\left(mol\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

0,1          0,1                           ( mol )

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

   0,1             0,3                                  ( mol )

\(\Rightarrow a=n_{H_2SO_4}=0,1+0,3=0,4\left(mol\right)\)

tỉ lệ mol của Fe0 và Fe2O3 là 1:1( mình nhấn nhầm )

Gọi \(\left\{{}\begin{matrix}n_{FeO}=n_{Fe_2O_3}=x\left(mol\right)\\n_{Fe_3O_4}=y\left(mol\right)\end{matrix}\right.\)

`=> 72x + 160x + 232y = 23,2`

`=> x + y = 0,1`

PTHH: \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\) (1)

            \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\) (2)

            \(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\) (3)

Theo PT: \(n_{H_2SO_4}=n_{FeO}+3n_{Fe_2O_3}+4n_{Fe_3O_4}=4a+4b\)

`=>` \(a=n_{H_2SO_4}=4.\left(a+b\right)=0,1.4=0,4\left(mol\right)\)

$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
$ZnO + H_2SO_4 \to ZnSO_4 + H_2O$
$MgO + H_2SO_4 \to MgSO_4 + H_2O$

Theo PTHH : $n_{H_2O} = n_{H_2SO_4} = 0,05(mol)$
Bảo toàn khối lượng : 

$m_{muối} = m_{hh} + m_{H_2SO_4} - m_{H_2O} = 2,81 + 0,05.98 - 0,05.18 = 6,81(gam)$

^{Fe2O3+ H2SO4--->FeSO4+H2O}

^{ZnO +H2SO4---> ZnSO4+ H2O}

^{MgO+ H2SO4---->MgSO4+H2O}

^{Do nH2SO4=0,05}

^{=>mH2SO4= 0,05.98=4,9 (g)}

^{Theo pt :nH2O=nH2SO4 =0,05 MOL}

^{mH2O=0,05.18=0,9 (g)}

^{ADDLBTKL ta có :2,81+4,9=m muối+0,9}

^{=> m muối=6,81(g)}

a)

$Na_2O + 2HCl \to 2NaCl + H_2O$
b)

$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$

Theo PTHH : 

$n_{NaCl} = 2n_{Na_2O} = 0,2(mol)$
$m_{NaCl} = 0,2.58,5 = 11,7(gam)$

c)

$n_{HCl} = 2n_{Na_2O} = 0,2(mol)$

$C_{M_{HCl}} = \dfrac{0,2}{0,3} = 0,67M$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Mg + H_2SO_4 \to MgSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$

Theo PTHH : $n_{H_2SO_4} = n_{H_2} = \dfrac{ \dfrac{5m}{67}}{2} = \dfrac{5m}{134}(mol)$

Bảo toàn khối lượng : 

$m_{hh} + m_{H_2SO_4} = m_{muối} + m_{H_2}$
$\Rightarrow m + \dfrac{5m}{134}.98 = 61,4 + \dfrac{5m}{67}$

$\Rightarrow m = 13,64(gam)$