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1, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\) (1:3:2:3)
2, \(2Na+2H_2O\rightarrow2NaOH+H_2\) (2:2:2:1)
3, \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\) (1:3:1:3)
4, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (2:3:1:3)
5, \(2Cu+O_2\underrightarrow{t^o}2CuO\) (2:1:2)
6, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (4:3:2)
7, \(FeO+2HCl\rightarrow FeCl_2+H_2O\) (1:2:1:1)
8, \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\) (1:2:1:1)
C1:
\(2KMnO_4\rightarrow K_2MnO_4+ MnO_2+O_2\)(tỉ lệ 2:1:1:1)
2Al(OH)\(_3\) + 3H\(_2\)SO\(_4\) → Al\(_2\)(SO4)\(_3\) + 6H2O(tỉ lệ 2:3:1:6)
\(4Na+O_2\rightarrow2Na_2O\)(tỉ lệ:4:1:2)
\(2Al+3Cl_2\rightarrow2AlCl_3\)(tỉ lệ:2:3:2)
\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)(tỉ lệ:2:1:3)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)(tỉ lệ 1:6:2:3)
\(4P+5O_2\rightarrow2P_2O_5\)(tỉ lệ:4:5:2)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)(tỉ lệ 1:3:1:3)
\(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)(tỉ lệ :1:2:1)
C2/
a,
\(mFeO=0,07.72=5,04g\)
\(mNa_2SO_4=0,25.142=35,5g\)
\(mK_2SO_4=0,03.174=5,22g\)
\(mH_2SO_4=0,25.98=24,5g\)
C3/
a,
\(VO_{2_{đkt}}=1,25.24=30lit\)
\(VO_{2_{đktc}}=1,25.22,4=28lit\)
b,
\(VN_{2_{\left(đkt\right)}}=0,125.24=3lit\)
\(VN_{2_{\left(đktc\right)}}=0,125.22,4=2,8lit\)
\(a,Mg+2HCl\rightarrow MgCl_2+H_2\\ 1:2:1:1\\ b,Fe_2O_3+3CO\rightarrow^{t^o}2Fe+3CO_2\uparrow\\ 1:3:2:3\\ c,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 2:3:1:3\\ d,2Al+3Cl_2\rightarrow^{t^o}2AlCl_3\\ 2:3:2\)
\(2Al+3Cl_2\rightarrow2AlCl_3\\ ....2.....3.....2\\ Sửa:Zn+2HCl\rightarrow ZnCl_2+H_2\\ ....1.....2.....1......1\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\\ ....2.....1.....3\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ ....2.....3.....1.....3\)
\(a.4Na+O_2-^{t^o}\rightarrow2Na_2O\left(4:1:2\right)\\ b.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\left(2:2:3\right)\\ c.FeO+2HCl\rightarrow FeCl_2+H_2\left(1:2:1:1\right)\\ d.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_2+3H_2O\left(1:3:1:3\right)\\ e.4P+5O_2-^{t^o}\rightarrow2P_2O_5\left(4:5:2\right)\)
a/ \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
b/ \(2Fe+3Cl_2\rightarrow2FeCl_3\)
c/ \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a. \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
tỉ lệ 2 : 3 : 1 : 3
b. \(2Fe+3Cl_2\rightarrow2FeCl_3\)
tỉ lệ 2 : 3 : 2
c. \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
tỉ lệ 1 : 6 : 2 : 3
\(a.2Al+3Cl_2\rightarrow2AlCl_3...2:3:2\\ b.Fe_3O_4+2C\rightarrow3Fe+2CO_2...1:2:3:2\\ c.2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O...2:3:1:6\)
Gọi số mol của Al2O3, Fe2O3 là 2a, 3a (mol)
PTHH: Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
2a---->6a
Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
3a----->9a
=> 6a + 9a = 0,3
=> a = 0,02
=> \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,04\left(mol\right)\\n_{Fe_2O_3}=0,06\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,04.102}{0,04.102+0,06.160}.100\%=29,825\%\\\%m_{Fe_2O_3}=\dfrac{0,06.160}{0,04.102+0,06.160}.100\%=70,175\%\end{matrix}\right.\)
Chọn D