PTHH: 3Fe + 2O₂ =(nhiệt)=> Fe₃O₄

a) nFe = 5,6 / 56 = 0,1 (mol)

\(\Rightarrow n_{O2}=\frac{0,1.2}{3}=\frac{1}{15}\left(mol\right)\)

Thể tích oxi cần dùng ở điều kiện tiêu chuẩn là:

=> V_O2(đktc) = \(\frac{1}{15}.22,4\approx1,5\left(lit\right)\)

b) n_Fe3O4 = \(\frac{0,1}{3}=\frac{1}{30}\left(mol\right)\)

=> m_Fe3O4 = \(\frac{1}{30}.232\approx7,73\left(gam\right)\)

\(3Fe+2O_2\rightarrow Fe_3O_4\)

\(m_{Fe_3O_4}=12+23.2=35.2\left(g\right)\)

a)\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(m\right)\)

\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

tỉ lệ        :2                  1                 1              1

số mol   :0,2               0,1              0,1           0,1

\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

b)\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(m\right)\)

\(PTHH:3Fe+2O_2\underrightarrow{ }Fe_3O_4\)

theo phương trình ta có tỉ lệ\(\dfrac{0,2}{3}>\dfrac{0,1}{2}\)=>Fe dư

\(PTHH:3Fe+2O_2\xrightarrow[]{}Fe_3O_4\)

tỉ lệ       :3           2         1

số mol  :0,15       0,1      0,05

\(m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\)

a. \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH : 3Fe + 2O₂ -t^o> 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 \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)

a, \(3Fe+2O_2\rightarrow Fe_3O_4\)

b, \(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

\(\Rightarrow3Fe+2O_2\rightarrow Fe_3O_{\text{4}}\)

     3    :   2      :   1

     0,3 :  0,2   : 0,1 

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

c,\(\Rightarrow3Fe+2O_2\rightarrow Fe_3O_{\text{4}}\)

        3    :   2      :   1

       0,3 :  0,2   : 0,1 

\(\Rightarrow V_{O_2}=0,2.22,4=4,48\left(lít\right)\)

\(n_{Fe}=\dfrac{6,8}{56}=0,12mol\)

3Fe + 2O₂ \(\underrightarrow{t^o}\) Fe₃O₄

0,12   0,08          0,04  ( mol )

a, \(V_{O_2}=0,08.22,4=1,792l\)

b, m_Fe3O4 = 0,04.232 = 9,28g

\(n_{Fe}=\dfrac{6,8}{56}=\dfrac{17}{140}(mol)\\ PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ a,n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{17}{210}(mol)\\ \Rightarrow V_{O_2}=\dfrac{17}{210}.22,4=1,81(g)\\ b,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{17}{420}(mol)\\ \Rightarrow m_{Fe_3O_4}=\dfrac{17}{420}.232=9,39(g)\)

\(a/3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ b/n_{Fe}=\dfrac{1,4}{56}=0,025mol\\ n_{O_2}=\dfrac{0,025.2}{3}=\dfrac{0,05}{3}mol\\ V_{O_2}=\dfrac{0,05}{3}\cdot22,4\approx0,37l\\ c/C_1\\ n_{Fe_3O_4}=\dfrac{0,025}{3}mol\\ m_{Fe_3O_4}=\dfrac{0,025}{3}\cdot232\approx1,93g\\ C_2\\ m_{O_2}=\dfrac{0,05}{3}\cdot32\approx0,53g\\ BTKL:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{Fe_3O_4}=1,4+0,53=1,93g\)

Ai giúp với

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

a, \(n_{Fe}=\dfrac{11}{56}\left(kmol\right)\)

Theo PT: \(n_{Fe_3O_4\left(LT\right)}=\dfrac{1}{3}n_{Fe}=\dfrac{11}{168}\left(kmol\right)\)

\(\Rightarrow m_{Fe_3O_4\left(LT\right)}=\dfrac{11}{168}.232=\dfrac{319}{21}\left(kg\right)\) > m_{Fe3O4 (TT)} = 200 (kg)

→ vô lý

Bạn xem lại đề phần a nhé.

b, \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(kmol\right)\)

Theo PT: \(n_{Fe\left(LT\right)}=3n_{Fe_3O_4}=0,3\left(kmol\right)\)

\(\Rightarrow m_{Fe\left(LT\right)}=0,3.56=16,8\left(kg\right)\)

Mà: H = 85%

\(\Rightarrow m_{Fe\left(TT\right)}=\dfrac{16,8}{85\%}=\dfrac{336}{17}\left(kg\right)\)

Theo gt ta có: $n_{Fe_3O_4}=0,05(mol)$

$Fe_3O_4+4H_2\rightarrow 3Fe+4H_2O$

Ta có: $n_{H_2}=0,05.4=0,2(mol)\Rightarrow V_{H_2}=4,48(l)$

 \(Fe3O4+4h2->3Fe+4H2O\)

a) n \(Fe3O4\\\)=\(\dfrac{11.6}{232}\)=0.05 mol

V\(H2=0,05.22,4=1,12l\)

b)n \(Fe=\dfrac{0.05\cdot3}{1}=0.15mol\)

 m\(Fe=n.M=0,15\cdot56=8,4gam\)