Đáp án:

Bài 1:  

$C\%_{HCl(dư)}=3,4\%$

$C\%_{FeCl_2}=2,5\%$

Bài 2: 

$C\%_{H_2SO_4(dư)}=8,6\%$

$C\%_{Al_2(SO_4)_3}=6,7\%$

Giải thích các bước giải:

  Bài 1:

     $FeO+2HCl→FeCl_2+H_2O$

Ta có: $n_{FeO}=\dfrac{2,16}{72}=0,03(mol)$

          $n_{HCl}=\dfrac{150×5\%}{36,5×100\%}≈0,2(mol)$

Xét $\dfrac{n_{FeO}}{1}$ và $\dfrac{n_{HCl}}{2}$

Ta thấy $n_{FeO}<n_{HCl}$

→ $FeO$ hết, $HCl$ dư.

 Tính theo số mol $FeO$

→ $n_{HCl(dư)}=0,2-0,03×2=0,14(mol)$

→ $n_{FeCl_2}=0,03(mol)$

Mặt khác: $m_{\text{dd sau pứ}}=2,16+150=152,16(g)$

⇒ $C\%_{HCl(dư)}=\dfrac{0,14×36,5}{152,16}×100\%=3,4\%$

⇒ $C\%_{FeCl_2}=\dfrac{0,03×127}{152,16}×100\%=2,5\%$

 Bài 2:

    $Al_2O_3+3H_2SO_4→Al_2(SO_4)_3+3H_2O$

Ta có: $n_{Al_2O_3}=\dfrac{2,04}{102}=0,02(mol)$

          $n_{H_2SO_4}=\dfrac{100×15\%}{98×100\%}≈0,15(mol)$

Xét $\dfrac{n_{Al_2O_3}}{1}$ và $\dfrac{n_{H_2SO_4}}{3}$

Ta thấy $n_{Al_2O_3}<n_{H_2SO_4}$

→ $Al_2O_3$ hết, $H_2SO_4$ dư.

 Tính theo số mol $Al_2O_3$

→ $n_{H_2SO_4(dư)}=0,15-0,02×3=0,09(mol)$

→ $n_{Al_2(SO_4)_3}=0,02(mol)$

Mặt khác: $m_{\text{dd sau pứ}}=2,04+100=102,04(g)$

⇒ $C\%_{H_2SO_4(dư)}=\dfrac{0,09×98}{102,04}×100\%=8,6\%$

⇒ $C\%_{Al_2(SO_4)_3}=\dfrac{0,02×342}{102,04}×100\%=6,7\%$

Đáp án:

`1)`

`C%_{dd\ FeCl_2}~~2,5%`

`C%_{dd\ HCl(du)}~~3,48%`

`2)`

`C%_{dd\ Al_2(SO_4)_3}~~6,7%`

`C%_{dd\ H_2SO_4(du)}~~8,93%`

Giải thích các bước giải:

`1)`

`Pt: FeO+2HCl->FeCl_2+H_2O`

`n_{FeO}={2,16}/{72}=0,03(mol)`

`n_{HCl}={150.5%}/{36,5.100%}=0,205(mol)`

`Ta\ co: {0,03}/1<{0,205}/2=>HCl\ du\ , FeO\ het`

`n_{HCl(pu)}=2n_{FeO}=0,06(mol)`

`n_{FeCl_2}=n_{FeO}=0,03(mol)`

`m_{dd\ spu}=2,16+150=152,16(g)`

`C%_{dd\ FeCl_2}={0,03.127}/{152,16}.100%~~2,5%`

`C%_{dd\ HCl(du)}={(0,205-0,06).36,5}/{152,16}.100%~~3,48%`

`2)`

`Pt: Al_2O_3+3H_2SO_4->Al_2(SO_4)_3+3H_2O`

`n_{Al_2O_3}={2,04}/{102}=0,02(mol)`

`n_{H_2SO_4}={100.15%}/{98.100%}=0,153(mol)`

`Ta\ co: {0,02}/1<{0,153}/3=>H_2SO_4\ du\ , Al_2O_3\ het`

`n_{H_2SO_4(pu)}=3n_{Al_2O_3}=0,06(mol)`

`n_{Al_2(SO_4)_3}=n_{Al_2O_3}=0,02(mol)`

`m_{dd\ spu}=2,04+100=102,04(g)`

`C%_{dd\ Al_2(SO_4)_3}={0,02.342}/{102,04}.100%~~6,7%`

`C%_{dd\ H_2SO_4(du)}={(0,153-0,06).98}/{102,04}.100%~~8,93%`