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a. \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH : 3Fe + 2O2 -to> Fe3O4
0,3 0,2 0,1
b. \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
c. \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
a \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b \(\Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3mol\) \(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1mol\Rightarrow m_{Fe_3O_4}=0,1\cdot232=2,32g\)
c \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)
\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{23,2}{232}=0,1mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,3 0,2 0,1 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,3.56=16,8g\)
\(V_{O_2}=n_{O_2}.22,4=0,2.22,4=4,48l\)
\(V_{kk}=\dfrac{4,48.100}{20}=22,4l\)
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,4 0,2 ( mol )
\(n_{KMnO_4}=\dfrac{0,4}{85\%}=\dfrac{8}{17}mol\)
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\dfrac{8}{17}.158=74,3529g\)
a)
\(b)n_{Fe_3O_4} = \dfrac{6,96}{232} = 0,03(mol)\\ 3Fe + 2O_2 \xrightarrow{t^o}Fe_3O_4\\ n_{Fe} = 3n_{Fe_3O_4} = 0,09(mol)\\ m_{Fe} = 0,09.56 = 5,04(gam)\\ c) n_{O_2} = 2n_{Fe_3O_4} = 0,06(mol)\\ V_{O_2} = 0,06.22,4 = 1,344(lít)\\ d) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,12(mol)\\ m_{KMnO_4} = 0,12.158 = 18,96(gam)\)
\(n_{Fe_3O_4}=\dfrac{6.96}{232}=0.03\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_2O_3\)
\(0.09.....0.06.......0.03\)
\(m_{Fe}=0.09\cdot56=5.04\left(g\right)\)
\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.12...............................................0.06\)
\(m_{KMnO_4}=0.12\cdot158=18.96\left(g\right)\)
n Fe3O4=\(\dfrac{13,92}{232}\)=0,06 mol
3Fe + 2O2 -to--> Fe3O4
0,18------0,12-------0,06
=>m Fe=0,18.56=10,08g
=>VO2=0,12.22,4=2,688l
2KMnO4-to>K2MnO4+MnO2+O2
0,24-------------------------------------0,12
=>m KMnO4=0,24.158=37,92g
nFe3O4 = 13,92 : 160= 0,087 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,087->0,058-->0,029 (mol)
=> mFe = 0,029 . 56 = 1,624 (g)
=> VO2 = 0,058 . 22,4 = 1,2992 (L)
pthh : 2KMnO4 -t--> K2MnO4 + MnO2 + O2
0,116<------------------------------0,058 (mol)
=> mKMnO4 = 0,116 . 158 = 18,328 (g)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\)
0,15 0,1 0,05
\(m_{Fe_2O_4}=0,05.232=11,6\left(g\right)\\
V_{O_2}=0,1.11,4=2,24\left(l\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,2 0,1
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe_3O_{\text{ 4}}}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(n_{O_2}=\dfrac{0.896}{22.4}=0.04\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(0.06......0.04.......0.02\)
\(m_{Fe}=0.06\cdot56=3.36\left(g\right)\)
\(m_{Fe_2O_3}=0.02\cdot232=4.64\left(g\right)\)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)
c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4\left(LT\right)}=2n_{O_2}=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow m_{KMnO_4\left(LT\right)}=\dfrac{2}{15}.158=\dfrac{316}{15}\left(g\right)\)
Mà: H% = 85%
\(\Rightarrow m_{KMnO_4\left(TT\right)}=\dfrac{\dfrac{316}{15}}{85\%}\approx24,78\left(g\right)\)
a.\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2mol\)
\(n_{O_2}=\dfrac{m}{M}=\dfrac{9,6}{32}=0,3mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
Xét: \(\dfrac{0,2}{4}\) < \(\dfrac{0,3}{5}\) ( mol )
0,2 0,1 ( mol )
\(m_{P_2O_5}=n.M=0,1.142=14,2g\)
b.\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{17,4}{232}=0,075mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,15 0,075 ( mol )
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,1 0,15 ( mol )
\(m_{KClO_3}=n.M=0,1.122,5=12,25g\)