Bài 2 :

`a,` PTHH:

`2A+2nHCl\to 2ACl_n+nH_2`

`2B+2nHCl\to 2BCl_n+nH_2`

`b,` `n_{H_2}=(9,916)/(24,79)=0,4(mol)`

Từ PTHH : `n_{HCl}=2n_{H_2}=0,4.2=0,8(mol)`

Bảo toàn khối lượng :

`a+36,5.0,8=67+0,4.2`

`=>a=35,6(g)`

Bài 3 :

`n_{BaCl_2}=(80.20,8\%)/(208)=0,08(mol)`

`n_{H_2SO_4}=(60.19,6\%)/(98)=0,12(mol)`

PTHH:

`BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl`

`0,08\to \,0,08\to 0,08\,\,\, \to 0,16(mol)`

`a,` `x=0,08.233=18,64(g)`

`b,` `m_{dd B}=80+60-18,64=121,36(g)`

`n_{H_2SO_4(\text{dư})}=0,12-0,08=0,04(mol)`

`C\%_{H_2SO_4}=(0,04.98)/(121,36).100\%≈3,23\%`

`C\%_{HCl}=(0,16.36,5)/(121,36).100\%≈4,81\%`

Bài 4 :

`n_{H_2}=(17,353)/(24,79)=0,7(mol)`

`a,` Gọi `n_{Zn}=a;n_{Fe}=b`

Ta có : `65a+56b=43,7(1)`

PTHH:

`Zn+2HCl\to ZnCl_2+H_2`

`a\to 2a`    `\to a`     `\to a(mol)`

`Fe+2HCl\to FeCl_2+H_2`

`b\to 2b`   `\to b`      `\to b(mol)`

`=>a+b=n_{H_2}=0,7(2)`

Từ (1) và (2):

`a=0,5;b=0,2`

`m_{Zn}=0,5.65=32,5(g)`

`m_{Fe}=0,2.56=11,2(g)`

`b,n_{HCl}=2n_{H_2}=2.0,7=1,4(mol)`

`=>m=(1,4.36,5)/(10\%)=511(g)`

`=>m_{dd(sau)}=43,7+511-0,7.2=553,3(g)`

`C\%_{ZnCl_2}=(136.0,5)/(553,3).100\%≈12,29\%`

`C\%_{FeCl_2}=(127.0,2)/(553,3).100\%≈4,59\%`