Bài 4 :

`n_{H_2}=(13,44)/(22,4)=0,6(mol)`

`n_{Fe_3O_4}=(23,2)/(232)=0,1(mol)`

PTHH:

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

Ta có tỉ lệ : `(0,1)/1 < (0,6)/4`

`=>` Sau phản ứng `Fe_3O_4` hết, `H_2` dư

`=>n_{Fe}= 1/3 n_{Fe_3O_4}= 1/3 . 0,1= 1/(30)(mol)`

`a=m_{Fe}= 1/(30).56≈1,87(g)`

PTHH:
`Fe+2HCl->FeCl_2+H_2`

`1/(30)`                    `->1/(30)(mol)`

`=>V= 1/(30).22,4≈0,747(lít)`

Bài 5 :

`CuO` không tan 

`=>m_{CuO}=8(g)`

`=>m_{Na_2O}=21,95-8=13,95(g)`

`=>n_{Na_2O}=(13,95)/(62)=0,225(mol)`

`1,`

`\%m_{CuO}=(8)/(21,95).100\%≈36,45\%`

`\%m_{Na_2O}=(100-36,45)\%=63,55\%`

`2,`

PTHH:

`Na_2O+H_2O->2NaOH`

`0,225`                `->0,45(mol)`

`m_{dd sau pứ}=100+13,95=113,95(g)`

`C\%_{NaOH}=(0,45.40)/(113,95).100\%≈13,8\%`

Bài 6 :

`n_{H_2}=(672:1000)/(22,4)=0,03(mol)`

Gọi `n_{Mg}=a;n_{Zn}=b`

Ta có : `24a+65b=1,54(1)`

`a,`PTHH:

`Mg+2HCl->MgCl_2+H_2`

`a->2a`     `->a`        `->a`

`Zn+2HCl->ZnCl_2+H_2`

`b`   `->2b`  `->b`     `->b`

ta có : `a+b=0,03(2)`

Từ (1) và (2) ta suy ra :

`a=0,01;b=0,02`

`b,`

`\%m_{Mg}=(0,01.24)/(1,54).100\%≈15,58\%`

`\%m_{Zn}=(100-15,58)\%=84,42\%`

`d,`

`m_{dd HCl}=(2a+2b).36,5:20\%=10,95(g)`

`c,`

`m_{dd sau pứ}=1,54+10,95-(0,01+0,02).2=12,43(g)`

`C\%_{MgCl_2}=(0,01.95)/(12,43).100\%≈7,64\%`

`C\%_{ZnCl_2}=(0,02.136)/(12,43).100\%≈21,88\%`