hoàn thành chuỗi
al(oh)3 -< al2(so4)3 -<alcl3-< al(no3)3-< al(oh)3-<al2o3-<alcl3
fe(oh)2 -< feso4-<fecl2_<fe(no3)2-<fe(oh)2, feo _<fecl2
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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.
\(\left(1\right)4Al+3O_2\overset{t^o}{--->}2Al_2O_3\)
\(\left(2\right)Al_2O_3+6HCl--->2AlCl_3+3H_2O\)
\(\left(3\right)AlCl_3+3NaOH--->Al\left(OH\right)_3+3NaCl\)
\(\left(4\right)2Al\left(OH\right)_3\overset{t^o}{--->}Al_2O_3+3H_2O\)
\(\left(5\right)Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2O\)
\(b,\left(1\right)4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \left(2\right)Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \left(3\right)AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ \left(4\right)2Al\left(OH\right)_3\rightarrow^{t^o}Al_2O_3+3H_2O\\ \left(5\right)Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(a,\left(1\right)Fe+2HCl\rightarrow FeCl_2+H_2\\ \left(2\right)FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl_2\\ \left(3\right)4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\\ \left(4\right)2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\\ \left(5\right)Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(a,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ 2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ Fe_2(SO_4)_3+3Zn\to 3ZnSO_4+2Fe\\ Fe+2HCl\to FeCl_2+H_2\\ b,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ Al_2O_3\xrightarrow[cryolit]{đpnc}4Al+3O_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\)
b) \(Al\underrightarrow{1}Al_2O_3\underrightarrow{2}Al\underrightarrow{3}Al_2\left(SO_4\right)_3\underrightarrow{4}Al\left(OH\right)_3\underrightarrow{5}Al_2O_3\)
(1) \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
(2) \(2Al_2O_3\xrightarrow[điện.phân.nóng.chảy]{criolit}4Al+3O_2\)
(3) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
(4) \(Al_2\left(SO_4\right)_3+6KOH\rightarrow2Al\left(OH\right)_3+3K_2SO_4\)
(5) \(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
Chúc bạn học tốt
Câu 3 :
200ml = 0,2l
\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,03 0,03 0,03
b) \(n_{H2}=\dfrac{0,03.1}{1}=0,03\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,03.24,79=0,7437\left(l\right)\)
c) \(n_{ZnCl2}=\dfrac{0,03.1}{1}=0,03\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,03}{0,2}=0,15\left(M\right)\)
Chúc bạn học tốt
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
\(2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+3BaCl_2\to 3BaSO_4\downarrow+2AlCl_3\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ 2Al(OH)_3\xrightarrow{t^o} Al_2O_3+3H_2O\)
1)
a) Từ trái qua phải :
\(4FeS+7O_2\xrightarrow[]{t^o}2Fe_2O_3+4SO_2\)
\(2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(Cu+2H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+H_2O+SO_2\uparrow\)
Bạn xem lại chỗ H2SO4 cho ra Cu nhé
b) Từ trái qua phải :
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Al_2\left(SO_4\right)_3+3NaOH\rightarrow Al\left(OH\right)_3+3Na_2SO_4\)
\(2Al\left(OH\right)_3\xrightarrow[]{t^o}Al_2O_3+3H_2O\)
2) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,2 -->0,6--------->0,2------>0,3
a) \(m_{ddHCl}=\dfrac{\left(0,6.36,5\right)}{7,3\%}.100\%=300\left(g\right)\)
b) \(m_{ddspu}=5,4+300-0,3.2=304,8\left(g\right)\)
\(C\%_{AlCl3}=\dfrac{0,2.133,5}{304,8}.100\%=8,75\%\)
a: \(Cu+H_2SO_4->CuSO_4+H_2\)
\(CuSO_4+BaCl_2->CuCl_2+BaSO_4\)
\(2CuCl_2+O_2->2CuO+2Cl_2\)
Al(OH)3 +H2SO4 --> Al2(SO4)3 + BaCl2 --> AlCl3 + AgNO3 --> Al(NO3)3 + Ba(OH)2 --> Al(OH)3 -to-> Al2O3 + HCl --> AlCl3
Fe(OH2) + H2SO4 --> FeSO4 + BaCl2 --> FeCl2 + AgNO3 --> Fe(NO3)2 + Ba(OH)2 --> Fe(OH)2 +O2 -to-> FeO +HCl--> FeCl2