$\text{Đáp án + giải thích các bước giải :}$

$Gọi$ $CTTQ$ $của$ $oxit$ $sắt$ $là$ $Fe_xO_y$ 

$PTHH$ $:$

$Fe_xO_y$ $+$ $2yHCl$ $→$ $xFeCl_{\frac{2y}{x}}$ $+$ $yH_2O$ $(1)$

$Fe$ $+$ $2HCl$ $→$ $FeCl_2$ $+$ $H_2↑$ $(2)$

$a/$

 $n_{H_2}$ $=$ $\frac{0,224}{22,4}$ $=$ $0,01$ $(mol)$

$Theo$ $(2)$ $:$ $n_{Fe}$ $=$ $n_{H_2}$ $=$ $0,01$ $(mol)$

$⇒$ $m_{Fe}$ $=$ $0,01$ $×$ $56$ $=$ $0,56$ $(g)$

$⇒$ $\%_{Fe}$ $=$ $\frac{0,56}{1,28}$ $×$ $100\%$ $=$ $43,75\%$ 

$⇒$ $\%_{Fe_xO_y}$ $=$ $100\%$ $-$ $43,75\%$ $=$ $56,25\%$ 

$b,$

$PTHH$ $:$

$Fe_xO_y$ $+$ $yH_2$ $\xrightarrow{t^o}$ $xFe$ $+$ $yH_2O$ $(3)$

$\%_{Fe_xO_y}$ $=$ $56,25\%$ 

$⇒$ $m_{Fe_xO_y}$ $=$ $6,4$ $×$ $56,25\%$ $=$ $3,6$ $(g)$

$m_O$ $=$ $6,4$ $-$ $5,6$ $=$ $0,8$ $(g)$

$⇒$ $n_O$ $=$ $\frac{0,8}{16}$ $=$ $0,05$ $(mol)$

$⇒$ $n_{Fe_xO_y}$ $=$ $\frac{0,05}{y}$ $(mol)$

$⇒$ $M_{Fe_xO_y}$ $=$ $\frac{3,6}{\frac{0,05}{y}}$ $=$ $72y$

$⇒$ $56x$ $+$ $16y$ $=$ $72y$ 

$⇒$ $x$ $=$ $y$ $=$ $1$

$⇒$ $CTHH$ $của$ $oxit$ $sắt$ $là$ $FeO$

$\text{Chúc bn hk tốt !! :))))}$

Đáp án:

`%m_{Fe}=43,75%\ ; %m_{Fe_xO_y}=56,25%`

`O x it: FeO`

Giải thích các bước giải:

`O x it : Fe_xO_y`

`Pt : Fe+2HCl->FeCl_2+H_2\ (1)`

       `Fe_xO_y+2yHCl->xFeCl_{{2y}/{x}}+yH_2O\ (2)`

       `Fe_xO_y+yH_2` $\xrightarrow{t^o}$ `xFe+yH_2O\ (3)`

`a)`

`n_{H_2(pt\ 1)}={0,224}/{22,4}=0,01(mol)`

`Theo\ pt\ (1) : n_{Fe}=n_{H_2}=0,01(mol)`

`m_{Fe}=0,01.56=0,56(g)`

`m_{o x it}=1,28-0,56=0,72(g)`

`%m_{Fe}={0,56}/{1,28}.100%=43,75%`

`%m_{Fe_xO_y}=100%-43,75%=56,25%`

`b)`

`1,28(g)\ hh\ A\ co\ 0,56(g)\ Fe`

`6,4(g)\ hh\ A\ co\ x(g)\ Fe`

`=>x={6,4.0,56}/{1,28}=2,8(g)`

`m_{o x it\ trong\ 6,4g\ hh\ A}=6,4-2,8=3,6(g)`

`Ta\ co: {3,6}/{56x+16y}={2,8}/{56x}`

`<=>3,6.56x=2,8.(56x+16y)`

`<=>201,6x=156,8x+44,8y`

`<=>44,8x=44,8y`

`=>x=y=1`

`O x it: FeO`