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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)
a, \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232=\dfrac{232}{15}\left(g\right)\)
c, \(n_{H_2}=\dfrac{4}{3}n_{Fe}=\dfrac{4}{15}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\)
d, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{Zn}=n_{H_2}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Zn}=\dfrac{4}{15}.65=\dfrac{52}{3}\left(g\right)\)
\(n_{HCl}=2n_{H_2}=\dfrac{8}{15}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{8}{15}.36,5=\dfrac{292}{15}\left(g\right)\)
\(a/3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ b/n_{Fe}=\dfrac{1,4}{56}=0,025mol\\ n_{O_2}=\dfrac{0,025.2}{3}=\dfrac{0,05}{3}mol\\ V_{O_2}=\dfrac{0,05}{3}\cdot22,4\approx0,37l\\ c/C_1\\ n_{Fe_3O_4}=\dfrac{0,025}{3}mol\\ m_{Fe_3O_4}=\dfrac{0,025}{3}\cdot232\approx1,93g\\ C_2\\ m_{O_2}=\dfrac{0,05}{3}\cdot32\approx0,53g\\ BTKL:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{Fe_3O_4}=1,4+0,53=1,93g\)
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 \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)
2)
nH2 = 6.72/22.4 = 0.3 (mol)
Fe2O3 + 3H2 -to-> 2Fe + 3H2O
0.1______0.3______0.2
mFe2O3 = 0.1*160 = 16 (g)
mFe = 0.2*56 = 11.2 (g)
3)
nFe3O4 = 11.6/232 = 0.05 (mol)
3Fe + 2O2 -to-> Fe3O4
0.15___0.1______0.05
mFe = 0.15*56 = 8.4 (g)
VO2 = 0.1*22.4 = 2.24 (l)
2KClO3 -to-> 2KCl + 3O2
1/15______________0.1
mKClO3 = 1/15 * 122.5 = 8.167 (g)
a)
3H2 + Fe2O3 --to--> 2Fe + 3H2O
b) nH2 = 6,72/22,4 = 0,3 mol
Từ pt => nFe3O4 = 0,1 mol
=> mFe3O4 = 0,1. 232 = 23,2 g
nFe = 33,6 : 56 = 0,6 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,6--> 0,4------->0,2 (mol)
=> vO2 = 0,4.22,4 = 8,96 (mol)
=> mFe3O4 = 0,2.232 = 46,4 (g)
pthh : 2KClO3 -t--> 2KClO3 + 3O2
0,267<-----------------------0,4(mol)
mKClO3= 0,267 .122,5 = 32,67 (g)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,3\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KClO_3}=0,4.122,5=49\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
\(a.PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 2 1
0,9 0,6 0,3
\(b.V_{O_2}=n.24,79=0,6.24,79=14,874\left(l\right)\)
\(c.m_{Fe_3O_4}=n.M=0,3.\left(56.3+16.4\right)=69,6\left(g\right)\)
\(d.V_{O_2}=14,874\left(l\right)\\ \Rightarrow n_{O_2}=\dfrac{V}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2 2 3
0,6 0,6 0,9
\(m_{KClO_3}=n.M=0,6.\left(39+35,5+16.3\right)=55,5\left(g\right).\)
a) \(3Fe+2O_2-^{t^o}\rightarrow Fe_3O_4\)
b) \(n_{Fe}=0,3\left(mol\right);n_{O_2}=0,2\left(mol\right)\)
Lập tỉ lệ : \(\dfrac{0,3}{3}=\dfrac{0,2}{2}\) => Cả 2 chất đều phản ứng hết
=> \(m_{O_2}=0,2.32=6,4\left(g\right)\)
c)Cách 1 :
Bảo toàn khối lượng => \(m_{Fe_3O_4}=m_{Fe}+m_{O_2}=16,8+6,4=23,2\left(g\right)\)
Cách 2: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1\left(mol\right)\)
=> \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
a) $3Fe +2 O_2 \xrightarrow{t^o} Fe_3O_4$
b) $n_{O_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
Theo PTHH : $n_{Fe} = \dfrac{3}{2}n_{O_2} = 0,3(mol)$
$m_{Fe} = 0,3.56 = 16,8(gam)$
c) $n_{Fe_3O_4} = \dfrac{1}{2}n_{O_2} = 0,1(mol)$
$m_{Fe_3O_4} = 0,1.232 = 23,2(gam)$