Đáp án:

 

Giải thích các bước giải:

Đặt CT $A_3:MCl_x$

$MCl_x+xAgNO_3\to M(NO_3)_x+xAgCl\downarrow$

$n_{AgCl}=\dfrac{14,35}{143,5}=0,1(mol)$

Theo PT: $n_{MCl_x}.x=n_{AgNO_3}=0,1$

$\Rightarrow \dfrac{6,35x}{M+35,5x}=0,1$

$\Rightarrow M+35,5x=63,5x$

$\Rightarrow M=28x$

Với $x=2\Rightarrow M=56(Fe)$

$\Rightarrow A_3:FeCl_2$

$A_1:Fe$

$A_2:HCl$

$A_3:FeCl_2$

$A_4:H_2$

$A_5:NaOH$

$A_6:Fe(OH)_2$

$A_7:NaCl$

$A_8:H_2O$

$A_9:O_2$

$A_{10}:Fe(OH)_3$

$A_{11}:Fe_2O_3$

$Fe+2HCl\to FeCl_2+H_2\uparrow$

$FeCl_2+2NaOH\to Fe(OH)_2\downarrow+2NaCl$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$4Fe(OH)_2+O_2+2H_2O\xrightarrow{t^o}4Fe(OH)_3\downarrow$

$2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O$

Giải thích các bước giải:

Gọi `XCl_n` là công thức tổng quát của `A_3`

`PTHH:`

`XCl_n + nAgNO_3 -> nAgCl + X(NO_3)_n`

`n_{AgCl_2} = {14,35}/{143,5} = 0,1(mol)`

Theo pt : `n_{XCl_2}.x = n_{AgCl} = 0,1(mol)`

`<=> {6,35.x}/{X + 35,5.n} = 0,1`

`<=> M = 28.n`

Biên luận 

\begin{array}{|c|c|c|}\hline \text{n}&\text{1}&\text{2}&\text{3}\\\hline M\text{}&\text{28}&\text{56(tm)}&\text{84}\\\hline\end{array}

CTHH `A_3` : `FeCl_2`

`A_1 : Fe`

`A_2 : HCl`

`A_3 : FeCl_2`

`A_4 : H_2`

`A_5 : NaOH`

`A_6 : Fe(OH)_2`

`A_7 : NaCl`

`A_8 : H_2O`

`A_9 : O_2`

`A_{10} : Fe(OH)_3`

`A_{11} : Fe_2O_3`

`PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2↑`

`FeCl_2 + NaOH -> Fe(OH)_2↓ + NaCl`

`Fe_2O_3 + 3H_2 ->^{t^0} 2Fe + 3H_2O`

`Fe(OH)_2 + O_2 + 2H_2O ->^{t^0} Fe(OH)_3↓`

`2Fe(OH)_3 ->^{t^0} Fe_2O_3 + 3H_2O`