Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2Al+3O_2\underrightarrow{^{^{t^o}}}2Al_2O_3\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\)
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
\(2Al\left(OH\right)_3\underrightarrow{^{^{t^o}}}Al_2O_3+3H_2O\)
\(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)
\(a,2Al_2O_3\rightarrow\left(đpnc,criolit\right)4Al+3O_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ 2Al\left(OH\right)_3\rightarrow\left(t^o\right)Al_2O_3+3H_2O\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(b,2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\)
$(1) 2Al_2O_3 \xrightarrow{đpnc} 4Al + 3O_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$AlCl_3 + 3KOH \to Al(OH)_3 + 3KCl$
$(2) Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$
$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl$
$(3) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$
$Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O$
Dãy biến hóa 1: Al → AlCl3 → Al(OH)3 → Al2O3
Dãy biến hóa 2: AlCl3 → Al(OH)3 → Al2O3 → Al PTHH
dãy biến hóa 1: 1): 2Al + 6HCl → 2AlCl3 + 3H2
(2): AlCl3 + 3NaOH → Al(OH)3 + 3NaCl
(3): 2Al(OH)3 t0 → Al2O3 + 3H2O PTHH
dãy biến hóa 2:
(1): AlCl3 +3NaOH → Al(OH)3+ 3NaCl
(2): 2Al(OH)3 t0 → Al2O3 + 3H2O
(3): 2Al2O3 dpnc → 4Al + 3O2
a)
\(Al+\dfrac{3}{2}Cl_2\rightarrow AlCl_3\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ 2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\\ Al_2O_3\xrightarrow[criolic]{đpnc}2Al+\dfrac{3}{2}O_2\)
b)
\(Fe+\dfrac{3}{2}Cl_2\rightarrow FeCl_3\\ FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\\ 2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\\ Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+3BaCl_2\to 2AlCl_3+3BaSO_4\downarrow\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow +3NaCl\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ 2Al_2O_3\xrightarrow{đpnc}4Al+3O_2\)
4Al+3O2--->2Al2O3
Al2O3+6HCl---->2Alcl3+3H2O
AlCl3+3NaOH--->3NaCl+Al(OH)3
2Al(OH)3--->Al2O3+3H2O
Al2O3+3H2--->2Al+3H2O
AlCl3 
Al(OH)3
Al2O3
Al
\(1.a.Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ 2Al\left(OH\right)_3-^{t^o}\rightarrow Al_2O_3+3H_2O\\ b.Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\\ 2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\\ Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)
2. Cho các mẫu thử vào dung dịch NaOH dư
+ Tan, có khí thoát ra: Al
\(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)
+ Không hiện tượng: Fe, Ag
Cho 2 mẫu thử không hiện tượng trên vào dung dịch HCl
+ Tan, có khí thoát ra: Fe
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
+ Không hiện tượng : Ag
(1) \(4Al+3O_2\xrightarrow[]{t^\circ}2Al_2O_3\)
(2) \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
(3) \(AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\)
(4) \(Al\left(NO_3\right)_3+3LiOH\rightarrow Al\left(OH\right)_3\downarrow+3LiNO_3\)
(5) \(2Al\left(OH\right)_3\xrightarrow[]{t^\circ}Al_2O_3+3H_2O\uparrow\)
(6) \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
4Al+3O2-->2Al2O2
Al2O3+6HCl---->2AlCl3+3H2O
AlCl3+3NaOH---->Al(OH)3+3NaCl
2Al(OH)3-->Al2O3+3H2O
Al2O3+3H2--->2Al+3H2O
Al2 O3 ko bị H2 phân hủy nhé