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![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(PTHH:3Fe+2O_2\rightarrow Fe_3O_4\)
Theo PTHH : \(n_{Fe}=3n_{Fe3O4}=3.\dfrac{m}{M}=0,45\left(mol\right)\)
\(\Rightarrow m_{Fe}=25,2g\)
b, Theo PTHH : \(n_{O2}=n_{Fe3O4}=0,3\left(mol\right)\)
\(\Rightarrow V_{O2}=6,72\left(l\right)\)
c, PTHH : \(2KClO_3\rightarrow2KCl+3O_2\)
Theo PTHH : \(n_{KClO3}=\dfrac{2}{3}n_{O2}=0,25\left(mol\right)\)
\(\Rightarrow m_{KClo3}=30,625g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(n_{O2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Pt : \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4|\)
3 2 1
0,3 0,15 0,075
Lập tỉ số so sánh : \(\dfrac{0,3}{3}>\dfrac{0,15}{2}\)
⇒ Fe dư , O2 phản ứng hết
⇒ Tính toán dựa vào số mol của O2
\(n_{Fe3O4}=\dfrac{0,15.1}{2}=0,075\left(mol\right)\)
⇒ \(m_{Fe3O4}=0,075.232=17,4\left(g\right)\)
\(n_{Fe\left(dư\right)}=0,3-\left(\dfrac{0,15.3}{2}\right)=0,075\left(mol\right)\)
⇒ \(m_{Fe\left(dư\right)}=0,075.56=4,2\left(g\right)\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a, nO2 = \(\frac{V}{22,4}\) = \(\frac{1,12}{22,4}\) = 0,05 ( mol )
nFe = \(\frac{m}{M}\) = \(\frac{8,4}{56}\) = 0,15 ( mol )
PTHH: 3Fe + 2O2 t\(\rightarrow\) Fe3O4
Theo PTHH: 2nO2 = 3nFe
3nFe = 2 . 0,05 = 0,1 ( mol )
Vậy sau phản ứng chất còn dư là Fe và còn dư : nFe = 0,15 - 0,1 = 0,05 ( mol ) \(\Rightarrow\)mFe = nFe . MFe = 0,05 . 56 = 2,8 ( g )
b, Theo PTHH: \(n_{Fe_3O_4}=2n_{O_2}=0,1\left(mol\right)\Rightarrow m_{Fe_3O_4}=M_{Fe_3O_4}\times n_{Fe_3O_4}=232\times0,1=23,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nO2 = 1,12/22,4 = 0,05 (mol)
PTHH: 3Fe + 2O2 -> (t°) Fe3O4
Mol: 0,075 <--- 0,05 ---> 0,025
mFe = 0,075 . 56 = 4,2 (g)
mFe3O4 = 0,025 . 232 = 5,8 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
nFe = 16.8/56 = 0.3 (mol)
3Fe + 2O2 -to-> Fe3O4
0.3......0.2...........0.1
VO2 = 0.2*22.4 = 4.48 (l)
mFe3O4 = 0.1*232 = 23.2 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(b)n_{Fe_3O_4} = \dfrac{6,96}{232} = 0,03(mol)\\ 3Fe + 2O_2 \xrightarrow{t^o}Fe_3O_4\\ n_{Fe} = 3n_{Fe_3O_4} = 0,09(mol)\\ m_{Fe} = 0,09.56 = 5,04(gam)\\ c) n_{O_2} = 2n_{Fe_3O_4} = 0,06(mol)\\ V_{O_2} = 0,06.22,4 = 1,344(lít)\\ d) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,12(mol)\\ m_{KMnO_4} = 0,12.158 = 18,96(gam)\)
\(n_{Fe_3O_4}=\dfrac{6.96}{232}=0.03\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_2O_3\)
\(0.09.....0.06.......0.03\)
\(m_{Fe}=0.09\cdot56=5.04\left(g\right)\)
\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.12...............................................0.06\)
\(m_{KMnO_4}=0.12\cdot158=18.96\left(g\right)\)
\(a) n_{O_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{Fe} = \dfrac{3}{2}n_{O_2} = 0,15(mol)\\ m_{Fe} = 0,15.56 = 8,4(gam)\\ b) \%Fe = \dfrac{56.3}{56.3+16.4}.100\% = 72,41\% \%O = 100\% - 72,41\% = 27,59\%\\ c) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,2(mol)\\ m_{KMnO_4} = 0,2.158 = 31,6(gam)\)
\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.15.......0.1......0.05\)
\(m_{Fe_3O_4}=0.05\cdot232=11.6\left(g\right)\)
\(\%Fe=\dfrac{0.05\cdot3\cdot56}{11.6}\cdot100\%=72.41\%\)
\(\%O=10072.41=27.59\%\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.2...................................................0.1\)
\(m_{_{ }KMnO_4}=0.2\cdot158=31.6\left(g\right)\)