1,Viết phương trình phản ứng thực hiện dãy chuyển hóa sau :
a, Al -> Al2O3 -> Al2(SO4)3 -> AlCl3 -> Al(OH)3 -> Al2O3 -> 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
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 |
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 |
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) \(2Fe+6H2SO4\left(\text{đ}\text{ă}c\right)-^{t0}->Fe2\left(SO4\right)3+6H2O+3SO2\uparrow\)
\(\left(2\right)Fe2\left(SO4\right)3+3Ba\left(NO3\right)2->2Fe\left(NO3\right)3+3B\text{aS}O4\downarrow\)
\(\left(3\right)Fe\left(NO3\right)3+3NaOH->Fe\left(OH\right)3\downarrow+3NaNO3\)
\(\left(4\right)2Fe\left(OH\right)3-^{t0}->Fe2O3+3H2O\)
\(\left(5\right)Fe2O3+3H2-^{t0}->2Fe+3H2O\)
1.
4Al + 3O2\(\rightarrow\)2Al2O3
Al2O3 + 3H2SO4(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 + 6H2SO4(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
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 \(\left(I\right),\left(II\right),\left(III\right),\left(IV\right)\Rightarrow a=...;b=...;c=...;d=...\)\(.\)
\(\Rightarrow\%m_{Fe}\)
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)
\(\Rightarrow m_{Fe}=6,72\left(g\right)\)
\(\Rightarrow\%m_{Fe}=\dfrac{6,72.100}{16}=42\%\)