Question

1,Viết phương trình phản ứng thực hiện dãy chuyển hóa sau :

a, Al -> Al₂O₃ -> Al₂(SO₄)₃ -> AlCl₃ -> Al(OH)₃ -> Al₂O_{3 ->} NaAlO₂

b.Fe -> Fe₂(SO₄)_{3 ->} Fe(NO₃)₃ -> Fe(OH)₃ ->Fe₂O₃ -> Fe

2, Lấy 16g hỗn hợp A ở dạng bột gồm Mg ,Al , Fe , Zn đem trộn đều rồi chia làm 2 phần bằng nhau . Phần 1 đốt cháy hoàn toàn trong Oxi dư thu được 11,52g hỗn hợp Oxit . Phần 2 hòa tan hoàn toàn trong dung dịch H₂SO₄ loãng dư thấy thoát ra 4,48 lít khí H₂ ở đktc . Tính % khối lượng của Fe trong hỗn hợp ban đầu ?

HELP ME !!!! Pls

Answer 1

Bài 1 : a,

4Al

+

3O2

→

2Al2O3

Al2O3

+

3H2SO4

→

Al2(SO4)3

+

3H2O

Al2(SO4)3

+

ZnCl2

→

AlCl3

+

ZnSO4

AlCl3

+

3NaOH

→

Al(OH)3

+

3NaCl

2Al(OH)3

→

Al2O3

+

3H2O

Al2O3

+

2NaOH

→

H2O

+

2NaAlO2

Answer 2

Bài 1 : b,

2Fe

+

6H2SO4

→

Fe2(SO4)3

+

6H2O

+

3SO2

3Ba(NO3)2

+

Fe2(SO4)3

→

2Fe(NO3)3

+

3BaSO4

3NaOH

+

Fe(NO3)3

→

3NaNO3

+

Fe(OH)3

2Fe(OH)3

→

Fe2O3

+

3H2O

2Al

+

Fe2O3

→

Al2O3

+

2Fe

Answer 3

Bài 1 :

a) \(Al->Al2O3->Al2\left(SO4\right)3->AlCl3->Al\left(OH\right)3->Al2O3->N\text{aA}lO2\)

(1) \(4Al+3O2-^{t0}->2Al2O3\)

(2) \(Al2O3+3H2SO4->Al2\left(SO4\right)3+3H2O\)

\(\left(3\right)Al2\left(SO4\right)3+3BaCl2->2AlCl3+3B\text{aS}O4\downarrow\)

\(\left(4\right)AlCl3+3NaOH->Al\left(OH\right)3\downarrow+3NaCl\)

\(\left(5\right)2Al\left(OH\right)3-^{t0}->Al2O3+3H2O\)

(6) \(Al2O3+2NaOH->2N\text{aA}lO2+H2O\)

b) \(Fe->Fe2\left(SO4\right)3->Fe\left(NO3\right)3->Fe\left(OH\right)3->Fe2O3->Fe\)

(1) \(2Fe+6H2SO4\left(\text{đ}\text{ă}c\right)-^{t0}->Fe2\left(SO4\right)3+6H2O+3SO2\uparrow\)

\(\left(2\right)Fe2\left(SO4\right)3+3Ba\left(NO3\right)2->2Fe\left(NO3\right)3+3B\text{aS}O4\downarrow\)

\(\left(3\right)Fe\left(NO3\right)3+3NaOH->Fe\left(OH\right)3\downarrow+3NaNO3\)

\(\left(4\right)2Fe\left(OH\right)3-^{t0}->Fe2O3+3H2O\)

\(\left(5\right)Fe2O3+3H2-^{t0}->2Fe+3H2O\)

Answer 4

1.

4Al + 3O₂\(\rightarrow\)2Al₂O₃

Al₂O₃ + 3H₂SO_4(l)\(\rightarrow\)Al₂(SO₄)₃ + 3H₂O

Al₂(SO₄)₃ + 3BaCl₂ \(\rightarrow\)2AlCl₃ + 3BaSO₄

AlCl₃ + 3NaOH \(\rightarrow\)Al(OH)₃ + 3NaCl

2Al(OH)₃ \(\underrightarrow{t^o}\)Al₂O₃ + 3H₂O

Al₂O₃ + 2NaOH \(\rightarrow\)2NaAlO₂ + H₂O

b;

2Fe + 6H₂SO_4(đ)\(\underrightarrow{t^o}\)Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

Fe₂(SO₄)₃ + 3Ba(NO₃)₂ \(\rightarrow\)2Fe(NO₃)₃ + 3BaSO₄

Fe(NO₃)₃ + 3NaOH \(\rightarrow\)Fe(OH)₃ + 3NaNO₃

2Fe(OH)₃ \(\underrightarrow{t^o}\)Fe₂O₃ + 3H₂O

Fe₂O₃ + 3H₂ \(\rightarrow\)2Fe + 3H₂O

Answer 5

Rainbow,chỉ coi bài 2 em cái

Answer 6

Hỗn hợp A\(\left\{{}\begin{matrix}Mg:a\left(mol\right)\\Al:b\left(mol\right)\\Fe:c\left(mol\right)\\Zn:d\left(mol\right)\end{matrix}\right.\)\(\Rightarrow24a+27b+56c+65d=16\left(I\right)\)

Chia làm 2 phần bằng nhau

\(2Mg\left(0,5a\right)+O_2\left(0,25a\right)-t^o->2MgO\left(0,5a\right)\)

\(4Al\left(0,5b\right)+3O_2\left(0,375b\right)-t^o->2Al_2O_3\left(0,25b\right)\)

\(3Fe\left(0,5c\right)+2O_2\left(\dfrac{1}{3}c\right)-t^o->Fe_3O_4\left(\dfrac{1}{6}c\right)\)

\(2Zn\left(0,5d\right)+O_2\left(0,25d\right)-t^o->2ZnO\left(0,5d\right)\)

\(m_{Oxit}=11,52\left(g\right)\)

\(\Leftrightarrow20a+25,5b+\dfrac{116}{3}c+40,5d=11,52\left(II\right)\)

Bảo toàn khối lượng \(\Rightarrow m_{O_2}=3,52\left(g\right)\Rightarrow n_{O_2}=0,11\left(mol\right)\)

\(\Leftrightarrow0,25a+0,375b+\dfrac{1}{3}c+0,25d=0,11\left(III\right)\)

\(Mg\left(0,5a\right)\rightarrow H_2\left(0,5a\right)\)

\(Al\left(0,5b\right)\rightarrow\dfrac{3}{2}H_2\left(0,75b\right)\)

\(Fe\left(0,5c\right)\rightarrow H_2\left(0,5c\right)\)

\(Zn\left(0,5d\right)\rightarrow H_2\left(0,5d\right)\)

(Ghi pthh ra nhé)

\(n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow0,5a+0,75b+0,5c+0,5d=0,2\left(IV\right)\)

Giair \(\left(I\right),\left(II\right),\left(III\right),\left(IV\right)\Rightarrow a=...;b=...;c=...;d=...\)\(.\)

\(\Rightarrow\%m_{Fe}\)

Answer 7

Ai giúp bài 2 với

Answer 8

2,

16 gam hỗn hợp \(\left\{{}\begin{matrix}Al:a\left(mol\right)\\Fe:b\left(mol\right)\\Zn:c\left(mol\right)\end{matrix}\right.\)\(\Rightarrow27a+56b+65c=16\left(I\right)\)

Chia làm 2 phần bằng nhau.

\(4Al\left(0,5a\right)+3O_2-t^o->2Al_2O_3\left(0,25a\right)\)

\(3Fe\left(0,5b\right)+2O_2-t^o->Fe_3O_4\left(\dfrac{1}{6}b\right)\)

\(2Zn\left(0,5c\right)+O_2-t^o->2ZnO\left(0,5c\right)\)

11,52 gam hỗn hợp oxit \(\left\{{}\begin{matrix}Al_2O_3:0,25a\left(mol\right)\\Fe_3O_4:\dfrac{1}{6}b\left(mol\right)\\ZnO:0,5c\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow25,5a+\dfrac{116}{3}b+40,5c=11,52\left(II\right)\)

\(n_{H_2}=0,2\left(mol\right)\)

\(2Al\left(0,5a\right)+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(0,75a\right)\)

\(Fe\left(0,5b\right)+H_2SO_4\rightarrow FeSO_4+H_2\left(0,5b\right)\)

\(Zn\left(0,5c\right)+H_2SO_4\rightarrow ZnSO_4+H_2\left(0,5c\right)\)

\(\Rightarrow0,75a+0,5b+0,5c=0,2\left(III\right)\)

Giari (I), (II) và (III) => b = 0,12 (mol)

\(\Rightarrow m_{Fe}=6,72\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{6,72.100}{16}=42\%\)