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\(a/3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ b/n_{Fe}=\dfrac{1,4}{56}=0,025mol\\ n_{O_2}=\dfrac{0,025.2}{3}=\dfrac{0,05}{3}mol\\ V_{O_2}=\dfrac{0,05}{3}\cdot22,4\approx0,37l\\ c/C_1\\ n_{Fe_3O_4}=\dfrac{0,025}{3}mol\\ m_{Fe_3O_4}=\dfrac{0,025}{3}\cdot232\approx1,93g\\ C_2\\ m_{O_2}=\dfrac{0,05}{3}\cdot32\approx0,53g\\ BTKL:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{Fe_3O_4}=1,4+0,53=1,93g\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,2 2/15 1/15 ( mol )
\(V_{kk}=V_{O_2}.5=\left(\dfrac{2}{15}.22,4\right).5=14,93l\)
\(m_{Fe_3O_4}=\dfrac{1}{15}.232=15,46g\)
\(a,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,\text{Bảo toàn KL: }m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ c,m_{O_2}=m_{Fe_3O_4}-m_{Fe}=28,4-12,4=16(g)\)
nFe3O4 = 23.2/232 = 0.1 mol
3Fe + 2O2 -to-> Fe3O4
0.3____0.2_______0.1
mFe = 0.3*56 = 16.8 g
VO2 = 0.2*22.4 = 4.48 (l)
Vkk = 5VO2 = 22.4 (l)
a, \(3Fe+2O_2\rightarrow Fe_3O_4\)
b, \(n_{Fe}=0,3\left(mol\right)\)
- Theo PTHH : \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V=n.22,4=4,48\left(l\right)\)
c, C1 : \(TheoPTHH:n_{Fe3o4}=\dfrac{1}{3}n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe3o4}=n.M=23,2\left(g\right)\)
C2: Áp dụng ddlbtkl : \(m_s=m_t=m_{Fe}+m_{O_2}=16,8+6,4=23,2\left(g\right)\)
Ô Lộc được mở lại nic rồi à tưởng vẫn bị khóa cơ
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
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: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{15}\cdot122,5\approx8,17\left(g\right)\)
a. \(n_{Fe}=\dfrac{5.6}{56}=0,1\left(mol\right)\)
PTHH : 3Fe + 2O2 ----to----> Fe3O4
0,1 \(\dfrac{0.2}{3}\) \(\dfrac{0.1}{3}\)
b. \(m_{Fe_3O_4}=\dfrac{0.1}{3}.232=\dfrac{23.2}{3}\left(g\right)\)
c. \(V_{O_2}=\dfrac{0.2}{3}.22,4=\dfrac{4.48}{3}\left(l\right)\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(nFe=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow nFe_3O_4=\dfrac{1}{3}.nFe=\dfrac{1}{3}.0,1=0,03\left(mol\right)\)
\(mFe_3O_4=0,03.232=6,96\left(g\right)\)
\(nO_2=\dfrac{2}{3}.nFe=\dfrac{2}{3}.0,1=0,07\left(mol\right)\)
\(VO_2=0,07.22,4=1,568\left(lít\right)\)