Viết phương trình có giấy chuyển hóa sau : Al->Al2O3->Al2(SO4)3->AlCl3->Al(OH3)
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(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) \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
\(a,2Al_2O_3\rightarrow\left(đpnc,criolit\right)4Al+3O_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ 2Al\left(OH\right)_3\rightarrow\left(t^o\right)Al_2O_3+3H_2O\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(b,2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\)
dễ
a, Fe + HCl(dư) ---> FeCl3
FeCl3 + NaOH ---> Fe(OH)3 + naCl
Fe(OH)3 ---> (nhiệt phân) Fe2O3 + h2O
fe2o3 + h2------> fe + h2o
b, al2o3 + h2 -------> al + h2o
Al+ h2SO4 -----------> AL2(SO4)3 + H2
AL2(SO4)3 + BACL2 ----------> alcl3 + baso4
alcl3 + naoh ------> nacl + al(oh)3
TỰ CÂN BẰNG NHA
a,1 2Fe+3Cl2->2FeCl3
2 FeCl3+NaOH->Fe(OH)3+NaCl
3 2Fe(OH)3->Fe2O3+3H2O(nhiệt phân)
4 Fe2O3+3CO->2Fe+3CO2
b,1 2Al2O3->4Al+3O2( điện phân nóng chảy)
2 2Al+3H2SO4(loãng)->Al2(SO4)3+3H2
3 Al2(SO4)3+3BaCl2->2AlCl3+3BaSO4
4 AlCl3+3NaOH->Al(OH)3+3NaCl
tick nha
\(4Al+3O_2\underrightarrow{^{^{t^o}}}2Al_2O_3\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)
\(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)
\(4Al+3O_2--^{t^o}->2Al_2O_3\)
\(Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2O\)
\(Al_2\left(SO_4\right)_3+3Cu\left(OH\right)_2--->2Al\left(OH\right)_3+3CuSO_4\)
\(Al\left(OH\right)_3+3HCl---->AlCl_3+3H_2O\)
\(2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+3BaCl_2\to 2AlCl_3+3BaSO_4\downarrow\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow +3NaCl\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ 2Al_2O_3\xrightarrow{đpnc}4Al+3O_2\)
a)
(1) 2Al + 6HCl --> 2AlCl3 + 3H2
(2) AlCl3 + 3NaOH --> Al(OH)3\(\downarrow\) + 3NaCl
(3) 2Al(OH)3 + 3H2SO4 --> Al2(SO4)3 + 6H2O
(4) Al2(SO4)3+ 3Ba(NO3)2 --> 2Al(NO3)3 + 3BaSO4\(\downarrow\)
b)
(1) 2Na + 2H2O --> 2NaOH + H2
(2) 2NaOH + H2SO4 --> Na2SO4 + 2H2O
(3) Na2SO4 + BaCl2 --> BaSO4\(\downarrow\) + 2NaCl
(4) NaCl + AgNO3 --> NaNO3 + AgCl\(\downarrow\)
c)
(1) 2Mg + O2 --to--> 2MgO
(2) MgO + 2HCl --> MgCl2 + H2O
(3) MgCl2 + 2NaOH --> Mg(OH)2\(\downarrow\) + 2NaCl
(4) Mg(OH)2 + H2SO4 --> MgSO4 + 2H2O
a)\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(2AlCl_3+3Ba\left(OH\right)_2\rightarrow2Al\left(OH\right)_3+3BaCl_2\)
\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(Al_2\left(SO_4\right)_3+3Ba\left(NO_3\right)_2\rightarrow2Al\left(NO_3\right)_3+3BaSO_4\)
b)\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\)
\(NaCl+HNO_3\rightarrow NaNO_3+HCl\)
Câu c em tự luyện tập nhé!!!
4Al+3O2----to--->2Al2O3
Al2O3 +3H2SO4----->Al2(SO4)3+3H2O
Al2(SO4)3+3BaCl2------>2AlCl3+3BaSO4
2AlCl3+3Ca(OH)2---->2Al(OH)3+3CaCl2
Al(OH)3+3HNO3---->Al(NO3)3+3H2O
2Al(NO3)3+3Mg----->3Mg(NO3)2+2Al
2Al+3H2SO4------->Al2(SO4)3+3H2
Al2(SO4)3+3Ca(OH)2------>2Al(OH)3+3CaSO4
2Al(OH)3-----to----->Al2O3+3H2O
2Al2O3----\(\dfrac{t^o}{criolit}\)---->4Al+3O2
2Al+2H2O+2NaOH------>2NaAlO2+3H2
4Al+3O2\(\underrightarrow{t^o}\) 2Al2O3
Al2O3+H2SO4 —> Al2(SO4)3+3H2O
Al2(SO4)3+3BaCl2–>3BaSO4\(\downarrow\)+2AlCl3
AlCl3+3NaOH—>Al(OH)3\(\downarrow\)+3NaCl
Al(OH)3+HNO3 —>Al(NO3)3+NO2\(\uparrow\)+H2O
2Al(NO3)3+3Mg—>3Mg(NO3)2+2Al\(\downarrow\)
2Al+3H2SO4–>Al2(SO4)3+3H2\(\uparrow\)
Al2(SO4)3+6NaOH—>2Al(OH)3+3Na2SO4
2Al(OH)3\(\underrightarrow{t^o}\)Al2O3+3H2O
2Al2O3\(\xrightarrow[criolit]{đien.phan.nong.chay}\)
4Al+3O2
2Al+2NaOH+2H2O—>2NaAlO2+3H2
\(4Al+3O_2\rightarrow2Al_2O_3\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)