\(\begin{array}{l}
2)\\
a)\\
n_{Fe}=\frac{11,2}{56}=0,2(mol)\\
Fe_xO_y+yH_2\xrightarrow{t^o}xFe+yH_2O\\
Theo\,PT:\,n_{Fe_xO_y}=\frac{n_{Fe}}{x}=\frac{0,2}{x}\\
\to \frac{16}{56x+16y}=\frac{0,2}{x}\\
\to \frac{x}{y}=\frac{2}{3}\\
\to x=2;y=3\\
\to Oxit:\,Fe_2O_3\\
b)\\
Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O\\
Theo\,PT:\,n_{H_2}=\frac{3}{2}n_{Fe}=0,3(mol)\\
\to V_{H_2(dùng)}=\frac{0,3}{100\%-10\%}.22,4=7,467(l)\\
3)\\
a)\\
Dat\,KL\,la\,R\\
100ml=0,1l\\
n_{HCl}=0,1.4=0,4(mol)\\
R+2HCl\to RCl_2+H_2\\
Theo\,PT:\,n_R=\frac{1}{2}n_{HCl}=0,2(mol)\\
\to M_R=\frac{13}{0,2}=65(g/mol)\\
\to R:\,kẽm\,(Zn)\\
b)\\
Zn+2HCl\to ZnCl_2+H_2\\
Theo\,PT:\,n_{Zn}=n_{H_2}=0,2(mol)\\
\to V_{H_2}=0,2.22,4=4,48(l)\\
c)\\
Theo\,PT:\,n_{ZnCl_2}=n_{Zn}=0,2(mol)\\
\to C_{M\,ZnCl_2}=\frac{0,2}{0,1}=2M
\end{array}\)

Đáp án + Giải thích các bước giải:

 a)

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

`n_{Fe_xO_y}=m/M=(16)/(56x + 16y) (mol)`

`n_{Fe}=m/M=(11,2)/(56)= 0,2 (mol)`

Theo PT: `n_{Fe_xO_y}=1/xn_{Fe}=1/x xx 0,2= (0,2)/x`

`<=> (16)/(56x + 16y) = (0,2)/x`

`<=> 0,2(56x + 16y) = 16x`

`<=> 11,2x + 3,2y = 16x`

`<=> 3,2y = 16x - 11,2x = 4,8x`

`=> x/y=(3,2)/(4,8)=2/3`

`=> x = 2, y = 3`

`=>` CTHH của oxit sắt: `Fe_2O_3`

b)

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

Theo PT: `n_{H_2}=3/2n_{Fe}=3/2 xx 0,2= 0,3 (mol)`

`=> V_{H_2}=(0,3 xx 22,4)/(100 - 10) xx 100= 7,467 (l)`

Bài 3:

a)

Gọi `R` là kim loại hóa trị `II`

`R + 2HCl->RCl_2 + H_2`

`n_{R}=m/M=(13)/(M_{R}) (mol)`

`n_{HCl}=C_M xx V=4 xx 0,1= 0,4 (mol)`

Theo PT: `n_{R}=1/2n_{HCl}=1/2 xx 0,4= 0,2 (mol)`

`<=> (13)/(M_{R}) = 0,2`

`=> M_{R}=(13)/(0,2)= 65` $(g/mol)$

`=> R`: Kẽm (`Zn`)

b)

`Zn + 2HCl->ZnCl_2 + H_2`

Theo PT: `n_{H_2}=n_{Zn}= 0,2 (mol)`

`=> V_{H_2}=n xx 22,4=0,2 xx 22,4= 4,48 (l)`

c)

Theo PT: `n_{ZnCl_2}=n_{Zn}= 0,2 (mol)`

`=> C_{M_{ZnCl_2}}=n/V=(0,2)/(0,1)= 2 (M)`