viết PTHH hoàn thành dãy biến hóa sau: Al->Al2O3->Al(SO4)->AlCl3->Al(OH)3->Al2O3->Al/NaAlO2
viết pthh hoàn thành dãy chuyển hóa sau
a) al2o3 -> al -> alcl3 -> al(oh)3 -> al2o3 -> al2(so4)3
b) cu -> cuo -> cucl2 -> cu(oh)2 -> cuo
*giúp tớ vs ạ :<<
\(a,2Al_2O_3\rightarrow\left(đpnc,criolit\right)4Al+3O_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ 2Al\left(OH\right)_3\rightarrow\left(t^o\right)Al_2O_3+3H_2O\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(b,2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\)
Viết các PTHH hoàn thành sơ đồ biến đổi sau : Al--->Al2O3--->AlCl3--->Al(OH)3--->Al2O3--->Al
\((1)4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\ (2)Al_2O_3+6HCl\to 2AlCl_3+3H_2O\\ (3)AlCl_3+3NaOH\to Al(OH)_3\downarrow +3NaCl\\ (4)2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ (5)2Al_2O_3\xrightarrow{đpnc}4Al+3O_2\)
Hoàn thành dãy biến hóa sau:
Al ––> Al2(SO4)3 ––> AlCl3 ––> Al(OH)3 ––> Al2O3
Al+H2SO4 -> Al2(SO4)3+ H2 \(\uparrow\)
Al2(SO4)3 +HCl -> AlCl3+ H2SO4
AlCl3 +Ca(OH)2 -> Al(OH)3 +CaCl2
Al(OH)3 - to> Al2O3 +H20
\(2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+3BaCl_2\to 3BaSO_4\downarrow+2AlCl_3\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ 2Al(OH)_3\xrightarrow{t^o} Al_2O_3+3H_2O\)
Al+H2SO4 -> Al2(SO4)3+ H2
↑
Al2(SO4)3 +HCl -> AlCl3+ H2SO4
AlCl3 +Ca(OH)2 -> Al(OH)3 +CaCl2
Al(OH)3 - to> Al2O3 +H20
Hoàn thành chuỗi biến hóa sau:
a) Fe → FeCl2 → Fe(OH)2 → Fe(OH)3 → Fe2O3 → FeCl3
b) Al → Al2O3 → AlCl3 → Al(OH)3 → Al2O3 → Al2(SO4)3
a.
\(\left(1\right)Fe+2HCl--->FeCl_2+H_2\)
\(\left(2\right)FeCl_2+2NaOH--->2NaCl+Fe\left(OH\right)_2\)
\(\left(3\right)4Fe\left(OH\right)_2+O_2+2H_2O\overset{t^o}{--->}4Fe\left(OH\right)_3\)
\(\left(4\right)2Fe\left(OH\right)_3\overset{t^o}{--->}Fe_2O_3+3H_2O\)
\(\left(5\right)Fe_2O_3+6HCl--->2FeCl_3+3H_2O\)
b.
\(\left(1\right)4Al+3O_2\overset{t^o}{--->}2Al_2O_3\)
\(\left(2\right)Al_2O_3+6HCl--->2AlCl_3+3H_2O\)
\(\left(3\right)AlCl_3+3NaOH--->Al\left(OH\right)_3+3NaCl\)
\(\left(4\right)2Al\left(OH\right)_3\overset{t^o}{--->}Al_2O_3+3H_2O\)
\(\left(5\right)Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2O\)
Hoàn thành chuỗi biến hóa sau:
a) Fe → FeCl2 → Fe(OH)2 → Fe(OH)3 → Fe2O3 → FeCl3
b) Al → Al2O3 → AlCl3 → Al(OH)3 → Al2O3 → Al2(SO4)3
\(b,\left(1\right)4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \left(2\right)Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \left(3\right)AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ \left(4\right)2Al\left(OH\right)_3\rightarrow^{t^o}Al_2O_3+3H_2O\\ \left(5\right)Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(a,\left(1\right)Fe+2HCl\rightarrow FeCl_2+H_2\\ \left(2\right)FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl_2\\ \left(3\right)4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\\ \left(4\right)2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\\ \left(5\right)Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Viết PTHH thực hiện dãy biến hóa sau (ghi rõ điều kiện xảy ra nếu có) 1, Al2O3 → Al → AlCl3 →Al(OH)3 2, Fe2O3 → Fe → FeCl3 → Fe(OH)3 3, Fe → FeCl3 → Fe(OH)3 → Fe2O3 → Fe → Fe2(SO4)3
$(1) 2Al_2O_3 \xrightarrow{đpnc} 4Al + 3O_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$AlCl_3 + 3KOH \to Al(OH)_3 + 3KCl$
$(2) Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$
$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl$
$(3) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$
$Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O$
al → al2o3 → al2(so4)3 → alcl3 → al(oh)3 → al(no3)3 → al → al2(so4)3 → al(oh)3 → al2o3 → al →naalo2
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
\(\begin{array}{l} 4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ Al_2(SO_4)_3+3BaCl_2\to 3BaSO_4\downarrow+2AlCl_3\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ Al(OH)_3+3HNO_3\to Al(NO_3)_3+3H_2O\\ 2Al(NO_3)_3+3Mg\to 3Mg(NO_3)_2+2Al\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\uparrow\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4\\ 2Al(OH)_3\xrightarrow{t^o} Al_2O_3+3H_2O\\ 2Al_2O_3\xrightarrow{\text{đpnc, Criolit}} 4Al+3O_2\uparrow\\ 2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow\end{array}\)
Giúp e vs ạ hoá lớp 9
Hoàn thành dãy biến hoá sau
Alcl3----->al(OH)
Al(OH)3 ---->NaAlo2
Al2o3 ------>NaAlo2
AlCl3 +3 NaOH-->Al(OH)3 + 3NaCl
Al(OH)3 + NaOH----->NaAlO2+2H2O
Al2O3+ 2NaOH------>2NaAlO2+H2O
Alcl3----->al(OH)
AlCl3 + 3NaOH -> Al(OH)3 + 3NaCl
Al(OH)3 ---->NaAlo2
Al(OH)3 + NaOH -> NaAlO2 + 2H2O
Al2o3 ------>NaAlo2
Al2O3 + 2NaOH -> 2NaAlO2 + H2O
phương trình hóa học biểu diễn dãy chuyển hóa sau Al2->Al2O3-> AlCl3 ->Al(no3)3-> Al(OH)3-> Al2O3-> NaAlO2
(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) \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)