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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,18 <------------------------ 0,18
\(\rightarrow n_O=\dfrac{13,92-0,18.56}{16}=0,24\left(mol\right)\)
CTHH: FexOy
=> x : y = 0,18 : 0,24 = 3 : 4
CTHH Fe3O4
![](https://rs.olm.vn/images/avt/0.png?1311)
Oxit sắt : FexOy
\(CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\\ n_{CO_2} = n_{CaCO_3} =\dfrac{22,5}{100} = 0,225(mol)\\ Fe_xO_y + yCO \xrightarrow{t^o} xFe + yCO_2\\ n_{oxit} = \dfrac{n_{CO_2}}{y} = \dfrac{0,225}{y}(mol)\\ \Rightarrow \dfrac{0,225}{y}(56x + 16y) = 12\\ \Rightarrow \dfrac{x}{y} = \dfrac{2}{3}\)
Vậy CTHH của oxit : Fe2O3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(V_{H_2\left(đktc\right)}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
1 : 1 (mol)
0,18 : 0,18 (mol)
\(yCO+Fe_xO_y\rightarrow^{t^0}xFe+yCO_2\uparrow\)
1 : x (mol)
\(\dfrac{0,18}{x}\) 0,18 (mol)
\(M_{Fe_xO_y}=\dfrac{m}{n}=\dfrac{13,92}{\dfrac{0,18}{x}}=\dfrac{232}{3}x\)
\(\Rightarrow56x+16y=\dfrac{232}{3}x\)
\(\Rightarrow16y=\dfrac{64}{3}x\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{16}{\dfrac{64}{3}}=\dfrac{3}{4}\Rightarrow x=3;y=4\)
-Vậy CTHH của oxit sắt là Fe3O4
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi công thức của oxit sắt đó là: FexOy
\(Fe_xO_y\left(b\right)+yCO\rightarrow xFe\left(bx\right)+yCO_2\left(by\right)\)
\(Fe\left(0,075\right)+H_2SO_4\rightarrow FeSO_4+H_2\left(0,075\right)\)
\(CO_2\left(0,1\right)+Ca\left(OH\right)_2\rightarrow CaCO_3\left(0,1\right)+H_2O\)
\(n_{H_2}=\frac{1,68}{22,4}=0,075\left(mol\right)\)
\(n_{CaCO_3}=\frac{10}{100}=0,1\left(mol\right)\)
Gọi số mol của oxit sắt là b thì ta có:
\(\left\{\begin{matrix}bx=0,075\\by=0,1\end{matrix}\right.\)
\(\Rightarrow\frac{x}{y}=\frac{0,075}{0,1}=\frac{3}{4}\)
\(\Rightarrow\left\{\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy oxit sắt đó là: Fe3O4
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
CTHH: FexOy
\(n_{Fe_xO_y}=\dfrac{16}{56x+16y}\left(mol\right)\)
PTHH: FexOy + yCO --to--> xFe + yCO2
\(\dfrac{16}{56x+16y}\)--------->\(\dfrac{16x}{56x+16y}\)
=> \(\dfrac{16x}{56x+16y}.56=16-4,8=11,2\)
=> \(\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow Fe_2O_3\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3CO --to--> 2Fe + 3CO2
0,1------>0,3--------------->0,3
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,3----->0,3
=> \(m_{CaCO_3}=0,3.100=30\left(g\right)\)
b) nCO (thực tế) = 0,3.110% = 0,33(mol)
=> VCO = 0,33.22,4 = 7,392(l)
![](https://rs.olm.vn/images/avt/0.png?1311)
X gồm Fe và Cu. Với HCl:
nFe = nH2 = 0,04
=>nCu = (mX – mFe)/64 = 0,02
=> nCuO = nFexOy = 0,02
-> x = nFe/nFexOy = 2
; Oxit là Fe2O3.
Bảo toàn O: \(m_{O\left(oxit\right)}=m_{giảm}=4,8-3,52=1,28\left(g\right)\)
\(n_{O\left(oxit\right)}=\dfrac{1,28}{16}=0,08\left(mol\right)\\ n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,04 <------------------------ 0,02
\(m_{Cu}=3,52-0,04.56=1,28\left(g\right)\\ n_{O\left(CuO\right)}=n_{Cu}=\dfrac{1,28}{64}=0,02\left(mol\right)\\ n_{O\left(Fe_xO_y\right)}=0,08-0,02=0,06\left(mol\right)\)
CTHH: FexOy
=> x : y = 0,04 : 0,06 = 2 : 3
CTHH Fe2O3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{FeCl_2}=\dfrac{25.4}{127}=0.2\left(mol\right)\)
\(n_{H_2O}=\dfrac{5.4}{18}=0.3\left(mol\right)\)
\(Fe_xO_y+yH_2\underrightarrow{t^0}xFe+yH_2O\)
...........................\(x\) ..........\(y\)
...........................\(0.2\) ......\(0.3\)
\(\Rightarrow0.3x=0.2y\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{0.2}{0.3}=\dfrac{2}{3}\)
\(CT:Fe_2O_3\)
\(m_{Fe_2O_3}=0.2\cdot2\cdot160=64\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
CTHH: FexOy
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: FexOy + yH2 --to--> xFe + yH2O
\(\dfrac{0,2}{x}\)<---------------0,2
Fe + 2HCl --> FeCl2 + H2
0,2<-------------------0,2
=> \(M_{Fe_xO_y}=56x+16y=\dfrac{16}{\dfrac{0,2}{x}}=80x\)
=> \(\dfrac{x}{y}=\dfrac{2}{3}\) => CTHH: Fe2O3
PTHH: \(Fe_xO_y+yCO-t^o->xFe+yCO_2\)(1)
\(Fe+H_2SO_4-->FeSO_4+H_2\)(2)
\(CO_2+Ca\left(OH\right)_2-->CaCO_3+H_2O\)(3)
\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
Theo PTHH (2) \(n_{Fe}=n_{H_2}=0,075\left(mol\right)\)
Theo PTHH (3) \(n_{CO_2}=n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
=> 0,1x = 0,075y
=> x=3, y=4
Vậy CT sắt là Fe3O4
=> \(m_{Fe_3O_4}=0,1.3.232=69,6\left(g\right)\)