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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 + 2O2 Fe3O4
b) nFe = \(\dfrac{8,4}{56}\)= 0,15 mol
nFe3O4 = \(\dfrac{11,6}{232}\) = 0,05 mol
Ta thấy \(\dfrac{nFe}{3}\)= \(\dfrac{nFe_3O_4}{1}\)=> Fe phản ứng hết
<=> nO2 cần dùng = \(\dfrac{2nFe}{3}\)= 0,1 mol
<=> mO2 cần dùng = 0,1.32 = 3,2 gam
c) Oxi chiếm thể tích bằng 1/5 thể tích không khí.
Mà V O2 = 0,1.22,4 = 2,24 lít => V không khí = 2,24 . 5 = 11,2 lít
\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\
pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
2,25 1,5
=> \(V_{O_2}=1,5.22,4=33,6\left(L\right)\)
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
1 1,5
=> \(m_{KClO3}=122,5\left(g\right)\)
a) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
0,3--->0,2----->0,1
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)
c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)
\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
0,44<------------------------------------0,22
\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)
a) PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b) Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,1\left(mol\right)\) \(\Rightarrow V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\)
c) PTHH: \(2KClO_3\xrightarrow[t^o]{MnO_2}2KCl+3O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{15}\cdot122,5\approx8,17\left(g\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\)
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
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
a)
PTHH: 3Fe + 2O2 ____\(t^o\)____> Fe3O4 (1)
b) Ta có: nFe = \(\dfrac{25.2}{56}=0.45\left(mol\right)\)
Theo (1): n\(O_2\)= \(\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}0.45=0.3\left(mol\right)\)
=> \(V_{O_2\left(đktc\right)}=0.3\cdot22.4=6.72\left(l\right)\)
c) PTHH: 2KClO3 __\(t^o\)___> 2KCl + 3O2 (2)
-Muốn điều chế được lượng oxi dùng cho phản ứng trên thì \(n_{O_2\left(2\right)}=n_{O_2\left(1\right)}=0.3\left(mol\right)\)
Theo (2) \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}0.3=0.2\left(mol\right)\)
=> \(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ Mol:0,4\rightarrow\dfrac{4}{15}\rightarrow\dfrac{2}{15}\)
\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\rightarrow V_{kk}=\dfrac{448}{75}.5=\dfrac{448}{15}\left(l\right)\\m_{Fe_3O_4}=\dfrac{2}{15}.232=\dfrac{464}{15}\left(g\right)\end{matrix}\right.\)
2KClO3 --to--> 2KCl + 3O2
\(\dfrac{8}{45}\) \(\dfrac{4}{15}\)
\(m_{KClO_3}=\dfrac{8}{45}.122,5=\dfrac{196}{9}\left(g\right)\)
a, \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: 3Fe + 2O2 ---to---> Fe3O4
Mol: 0,4 \(\dfrac{0,8}{3}\) \(\dfrac{0,4}{3}\)
b, \(V_{O_2}=\dfrac{0,8}{3}.22,4=5,973\left(l\right)\)
c, \(V_{kk}=\dfrac{448}{75}.5=29,867\left(l\right)\)
d, \(m_{Fe_3O_4}=\dfrac{0,4}{3}.232=30,93\left(g\right)\)
e,
PTHH: 2KClO3 ---to---> 2KCl + 3O2
Mol: \(\dfrac{0,16}{9}\) \(\dfrac{0,8}{3}\)
\(m_{KClO_3}=\dfrac{0,16}{9}.122,5=2,178\left(g\right)\)