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a)\(n_{Fe_3O_4}=\dfrac{69,6}{232}=0,3\left(m\right)\)
\(PTHH:3Fe+2O_2\xrightarrow[]{}Fe_3O_4\)
tỉ lệ : 3mol 2mol 1mol
số mol : 0,9 0,6 0,3
\(m_{Fe}=0,9.56=50,4\left(g\right)\)
\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
b)\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
tỉ lệ : 2mol 2mol 3mol
số mol : 0,4 0,4 0,6
\(m_{KClO_3}=122,5.0,4=49\left(g\right)\)
\(n_{Fe_3O_4}=\dfrac{2.32}{232}=0.01\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{t^0}}Fe_3O_4\)
\(0.03......0.02.........0.01\)
\(m_{Fe}=0.03\cdot56=1.68\left(g\right)\)
\(m_{O_2}=0.02\cdot32=0.64\left(g\right)\)
\(2KMnO_4\underrightarrow{^{t^0}}K_2MnO_4+MnO_2+O_2\)
\(0.04............................................0.02\)
\(m_{KMnO_4}=0.04\cdot158=6.32\left(g\right)\)
a)
n Fe3O4 = 2,32/232 = 0,01(mol)
3Fe + 2O2 \(\xrightarrow{t^o}\) Fe3O4
0,03....0,02.......0,01...........(mol)
m Fe = 0,03.56 = 1,68(gam)
m O2 = 0,02.32= 0,64(gam)
c)
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
n KMnO4 = 2n O2 = 0,04(mol)
m KMnO4 = 0,04.158 = 6,32 gam
a, Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
THeo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\Rightarrow V_{O_2}=0,04.22,4=0,896\left(l\right)\)
b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,08\left(mol\right)\Rightarrow m_{KMnO_4}=0,08.158=12,64\left(g\right)\)
a) \(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02\left(mol\right)\).
PTHH : \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Mol : 3 : 2 : 1
Mol 0,04 ← 0,02
\(\Rightarrow V_{O_2}=n_{O_2}.22,4=\left(0,04\right).\left(22,4\right)=0,896\left(l\right)\).
b) Từ phương trình ở câu a \(\Rightarrow n_{O_2}=0,04\left(mol\right)\).
PTHH : \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Mol : 2 : 1 : 1 : 1
Mol : 0,08 ← 0,04
\(\Rightarrow m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\left(0,08\right).158=12,64\left(g\right)\).
a)\(n_{Fe_3O_4}=\dfrac{11,6}{232}=0,05mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe}=0,15\cdot56=8,4g\)
\(m_{O_2}=0,1\cdot32=3,2g\)
b)\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,2 0,1
\(m_{KMnO_4}=0,2\cdot158=31,6g\)
\(a,n_{Fe_3O_4}=\dfrac{11,6}{232}=0,05\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,15<--0,1<----------0,05
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,15.56=8,4\left(g\right)\\m_{O_2}=0,1.32=3,2\left(g\right)\end{matrix}\right.\)
b, PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,2<--------------------------------------0,1
=> mKMnO4 = 0,2.158 = 31,6 (g)
a.\(\%Fe=\dfrac{56.3}{56.3+16.4}.100=72,41\%\)
b.\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,04 0,02 ( mol )
\(m_{O_2}=0,04.32=1,28g\)
c.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,08 0,04 ( mol )
\(m_{KMnO_4}=0,08.158=12,64g\)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
\(m_{O_2}=0,04.32=1,28\left(g\right)\)
b, Phần này đề bài cho là KMnO4 hay KClO3 vậy bạn?
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{6,96}{56\cdot3+16\cdot4}=0,03\left(mol\right)\\ PTHH;3Fe+2O_2-^{t^o}>Fe_3O_4\)
tỉ lệ: 3 : 2 : 1
n(mol) 0,09<-----0,06<---0,03
\(m_{Fe}=n\cdot M=0,09\cdot56=5,04\left(g\right)\\ V_{O_2\left(dktc\right)}=n\cdot22,4=0,06\cdot22,4=1,344\left(l\right)\)
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,06 0,04 0,02 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)
\(V_{O_2}=n_{O_2}.22,4=0,04.22,4=0,896l\)
Sửa đề: 4,46 (g) → 4,64 (g)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo PT: \(n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
\(n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\Rightarrow m_{O_2}=0,04.32=1,28\left(g\right)\)
b, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,08\left(mol\right)\Rightarrow m_{KMnO_4}=0,08.158=12,64\left(g\right)\)
\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,06 0,04 0,02 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)
\(m_{O_2}=n_{O_2}.M_{O_2}=0,04.32=1,28g\)
\(pthh:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo pt: \(n_{Fe}=3.0,02=0,06\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
Theo pt: \(n_{O_2}=2.0,02=0,04\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,04.32=1,28\left(g\right)\)