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\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ FeCl_3+3KOH\rightarrow Fe\left(OH\right)_3+3KCl\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
2Fe+3Cl2→(to)2FeCl3
FeCl3+3KOH→Fe(OH)3+3KCl2
Fe(OH)3→(to)Fe2O3+3H2O
Fe2O3+6HCl→2FeCl3+3H2O
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.
\(\left(1\right)4Al+3O_2\overset{t^o}{--->}2Al_2O_3\)
\(\left(2\right)Al_2O_3+6HCl--->2AlCl_3+3H_2O\)
\(\left(3\right)AlCl_3+3NaOH--->Al\left(OH\right)_3+3NaCl\)
\(\left(4\right)2Al\left(OH\right)_3\overset{t^o}{--->}Al_2O_3+3H_2O\)
\(\left(5\right)Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2O\)
\(b,\left(1\right)4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \left(2\right)Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \left(3\right)AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\\ \left(4\right)2Al\left(OH\right)_3\rightarrow^{t^o}Al_2O_3+3H_2O\\ \left(5\right)Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(a,\left(1\right)Fe+2HCl\rightarrow FeCl_2+H_2\\ \left(2\right)FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl_2\\ \left(3\right)4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\\ \left(4\right)2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\\ \left(5\right)Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(Fe\underrightarrow{1}Fe_2\left(SO_4\right)_3\underrightarrow{2}Fe\left(OH\right)_3\underrightarrow{3}Fe_2O_3\underrightarrow{4}Fe\underrightarrow{5}FeCl_2\underrightarrow{6}Fe\left(NO_3\right)_2\)
(1) \(2Fe+6H_2SO_{4đặc}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
(2) \(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)
(3) \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
(4) \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
(5) \(Fe+2HCl\rightarrow FeCl_2+H_2\)
(6) \(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\)
Chúc bạn học tốt
\(a) 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ 2FeCl_3 + Fe \rightarrow 3FeCl_2\)
\(b) 6nCO_2 + 5nH_2O \xrightarrow[\text{chất diệp lục}]{\text{ánh sáng}} (-C_6H_{10}O_5-)_n + 6nO_2\\ (-C_6H_{10}O_5-)_n + nH_2O \xrightarrow{axit} nC_6H_{12}O_6\\ C_6H_{12}O_6 \xrightarrow{\text{men rượu}} 2CO_2 + 2C_2H_5OH\\ C_2H_5OH + O_2 \xrightarrow{\text{men giấm}} CH_3COOH + H_2O\)
\(a,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ 2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ Fe_2(SO_4)_3+3Zn\to 3ZnSO_4+2Fe\\ Fe+2HCl\to FeCl_2+H_2\\ b,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ Al_2O_3\xrightarrow[cryolit]{đpnc}4Al+3O_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\)
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
\(Fe\left(NO_3\right)_3+3KOH\rightarrow Fe\left(OH\right)_3+3KNO_3\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCl_2+2KOH\rightarrow Fe\left(OH\right)_2+2KCl\)
Sửa đề chất cuối thành Fe(OH)2