Đáp án:

Bài `1:m_{Fe}=32(g)`

Bài `2:V=6,72(l)``

Giải thích các bước giải:

Bài `1:`

`n_{H_2}=\frac{4,48}{22,4}=0,2(mol)`

Phương trình hóa học:

$FeO+H_2\xrightarrow[]{t^o}Fe↓+H_2O$

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

$Fe_3O_4+4H_2\xrightarrow[]{t^o}3Fe↓+4H_2O$

`n_{H_2O}=n_{H_2}=0,2(mol)`

$\xrightarrow[]{\text{BTKL}}m_{hh}+m_{H_2}=m_{Fe}+m_{H_2O}$

` \to 35,2+0,2.2=m_{Fe}+0,2.18`

` \to m_{Fe}=32(g)`

Bài `2:`

Gọi `n_{H_2}=x(mol)`

Phương trình hóa học:

$CuO+H_2\xrightarrow[]{t^o}Cu↓+H_2O$

$FeO+H_2\xrightarrow[]{t^o}Fe↓+H_2O$

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

$Fe_3O_4+4H_2\xrightarrow[]{t^o}3Fe↓+4H_2O$

`n_{H_2O}=n_{H_2}=x(mol)`

$\xrightarrow[]{\text{BTKL}}m_{hh}+m_{H_2}=m_{X}+m_{H_2O}$

` \to 44,8+2.x=40+18.x`

` \to x=0,3 \to n_{H_2}=0,3(mol)`

`V=V_{H_2}(đktc)=0,3.22,4=6,72(l)`

Đáp án+ Giải thích các bước giải:

Bài 1:

`n_{H_2}= V/{22,4}= {4,48}/{22,4} = 0,2 (mol)`

`PT : FeO + H_2` $\xrightarrow{t^o}$ `Fe + H_2O`

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

        `Fe_3O_4 + 4H_2` $\xrightarrow{t^o}$ `3Fe + 4H_2O`

BTNT H: `n_{H_2O} = n_{H_2} = 0,2 (mol)`

BTKL : `m_{cr} + m_{H_2} = m_{Fe} + m_{H_2O}`

`=>m_{Fe}= 35,2 + 0,2.2 - 0,2.18`

`=>m_{Fe} = 32 (g)`

Bài 2:

`PT : CuO + H_2` $\xrightarrow{t^o}$ `Cu + H_2O`

        `FeO + H_2` $\xrightarrow{t^o}$ `Fe + H_2O`

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

        `Fe_3O_4 + 4H_2` $\xrightarrow{t^o}$ `3Fe + 4H_2O`

Gọi `x(mol): n_{H_2}`

BTNT H : `n_{H_2O} = n_{H_2} = x (mol)`

BTKL : `m_{A} + m_{H_2} = m_{X} + m_{H_2O}`

`<=>44,8+2x=40+18x`

`=>x=0,3(mol)`

`V_{H_2}=nxx22,4=0,3xx22,4=6,72(l)=V`

$\text{#Cyrus}$