Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Al}=\frac{2,7}{27}=0,1(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

Theo PT: $n_{H_2}=\frac{3}{2}n_{Al}=0,15(mol)$

$\to V_{H_2}=0,15.22,4=3,36(l)$

$b\big)$

$n_{Al}=\frac{10,8}{27}=0,4(mol)$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

Theo PT: $n_{O_2}=\frac{3}{4}n_{Al}=0,3(mol)$

$\to V_{kk}=0,3.22,4.5=33,6(l)$

$c\big)$

$200ml=0,2l$

Theo PT: $n_{HCl}=2n_{H_2}=0,3(mol)$

$\to C_{M\,HCl}=\frac{0,3}{0,2}=1,5M$

Theo PT: $n_{AlCl_3}=n_{Al}=0,1(mol)$

$\to C_{M\,AlCl_3}=\frac{0,1}{0,2}=0,5M$

$d\big)$

Đặt oxit là $Fe_xO_y$

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

$n_{Fe}=\frac{5,6}{56}=0,1(mol)$

Theo PT: $\frac{n_{H_2}}{y}=\frac{n_{Fe}}{x}$

$\to\frac{0,15}{y}=\frac{0,1}{x}$

$\to\frac{x}{y}=\frac{2}{3}\to x=2;y=3$

Vậy oxit là $Fe_2O_3$

Câu 2 :

`n_{Al}=(2,7)/(27)=0,1(mol)`

`a,`

PTHH:

`2Al+6HCl->2AlCl_3+3H_2`

`0,2->0,6`   `->0,2`   `->0,15(mol)`

`V_{H_2}=0,3.22,4=6,72(lít)`

`b,`

`n_{Al}=(10,8)/(27)=0,4(mol)`

PTHH:
`4Al+3O_2`$\xrightarrow[]{t^o}$`2Al_2O_3`

`0,4->0,3(mol)`

`=>V_{O_2}=0,3.22,4=6,72(lít)`

`V_{O_2}= 1/5 V_{kk}`

`=>V_{kk}=5V_{O_2}=5.6,72=33,6(lít)`

`c,`

`C_{M HCl}=(0,6)/(200:1000)=3(M)`

`C_{M AlCl_3}=(0,2)/(200:1000)=1(M)`

`d,`

`n_{Fe}=(5,6)/(56)=0,1(mol)`

Gọi CTHH của oxit sắt đó là `Fe_xO_y`

PTHH:

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

                `(0,1y)/x`  `←0,1(mol)`

`=>n_{H_2}=(0,1y)/x =0,15`

`=> y/x=3/2`

`=>x:y= 2:3`

`=>x=2;y=3`

`->` Oxit sắt : `Fe_2O_3`