Question

Answer 1

- Xét phần 2:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2NaOH + 2Al + H₂O ---> 2NaAlO₂ + 3H₂

                             0,1<-----------------------------0,15

- Xét phần 1:

Gọi n_Fe = a (mol) (a > 0)

PTHH:

Fe + H₂SO₄ ---> FeSO₄ + H₂

a--------------------->a

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

0,1---------------------->0,05

FeSO₄ + 2NaOH ---> Fe(OH)₂↓ + Na₂SO₄

a---------------------------->a

Al₂(SO₄)₃ + 6NaOH ---> 2Al(OH)₃ + 3Na₂SO₄

0,05---------------------------->0,1

4Fe(OH)₂ + O₂ --t^o--> 2Fe₂O₃ + 2H₂O

a------------------------------>0,5a

2Al(OH)₃ --t^o--> Al₂O₃ + 3H₂O

0,1----------------->0,05

=> 160.0,5a + 0,05.102 = 8

=> a = 0,03625 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{0,03625.56+0,1.27}.100\%=57,08\%\\\%m_{Fe}=100\%-57,08\%=42,92\%\end{matrix}\right.\)