A

A

a

\(a,3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{16,8}{56}=0,3\left(kmol\right)\\ n_{O_2}=\dfrac{2}{3}.0,3=0,2\left(kmol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,2.1000.22,4=4480\left(l\right)\\ n_{Fe_3O_4}=\dfrac{1}{3}.0.3=0,1\left(kmol\right)\\ m_{Fe_3O_4}=232.0,1=23,2\left(kg\right)\)

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

\(Áp.dụng.ĐLBTKL,ta.có:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{O_2}=m_{Fe_3O_4}-m_{Fe}=23,2-16,8=6,4\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{m}{M}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ \Rightarrow V_{O_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

3Fe+2O2->Fe3O4

n_Fe3O4=23,2/232=0,1 mol

=>n_O2=0,1x2=0,2 mol

V_O2=0,2x22,4=4,48 l 

a)

\(b)n_{Fe_3O_4} = \dfrac{6,96}{232} = 0,03(mol)\\ 3Fe + 2O_2 \xrightarrow{t^o}Fe_3O_4\\ n_{Fe} = 3n_{Fe_3O_4} = 0,09(mol)\\ m_{Fe} = 0,09.56 = 5,04(gam)\\ c) n_{O_2} = 2n_{Fe_3O_4} = 0,06(mol)\\ V_{O_2} = 0,06.22,4 = 1,344(lít)\\ d) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,12(mol)\\ m_{KMnO_4} = 0,12.158 = 18,96(gam)\)

\(n_{Fe_3O_4}=\dfrac{6.96}{232}=0.03\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.09.....0.06.......0.03\)

\(m_{Fe}=0.09\cdot56=5.04\left(g\right)\)

\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(0.12...............................................0.06\)

\(m_{KMnO_4}=0.12\cdot158=18.96\left(g\right)\)

đến nhà mk thì mk mới giúp

 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)   

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

a)

\(2Cu + O_2 \xrightarrow{t^o} 2CuO\)

b)

\(n_{CuO} = n_{Cu} = \dfrac{6,4}{64} = 0,1(mol)\\ \Rightarrow m_{CuO} = 0,1.80 = 8(gam)\)

c)

\(n_{O_2} = \dfrac{1}{2}n_{Cu} = 0,05(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ m_{KMnO_4} = 2n_{O_2} = 0,05.2 = 0,1.158 = 15,8(gam)\)

d)

\(V_{không\ khí} = 5V_{O_2} = 0,05.22,4.5 = 5,6(lít)\)

a) \(n_{Fe}=\dfrac{3,36}{56}=0,06\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

          0,06->0,04------->0,02

=> m_Fe3O4 = 0,02.232 = 4,64 (g)

b) V_O2 = 0,04.22,4 = 0,896 (l)

\(n_{Fe}=\dfrac{33,6}{56}=0,6mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

 0,6     0,4                0,2     ( mol )

\(V_{kk}=0,4.22,4.5=44,8l\)

\(m_{Fe_3O_4}=0,2.232=46,4g\)

3Fe+2O2-to>Fe3O4

0,6-----0,4-------0,2 mol

n Fe=\(\dfrac{33,6}{56}\)=0,6 mol

=>Vkk=0,4.5.22,4=44,8l

m Fe3O4=0,2.232=46,4g

nFe = 33,6 / 56 = 0,6 mol
pthh: 3Fe + 2O2 --to-> Fe3O4
0,6 ----------> 0,4 -------> 0,2
a. VO2 = 0,4 * 22,4 = 8,96 lít
VO2 = 20% Vkk => Vkk = 8,96 * 100 / 20 = 44,8 lít
b. mFe3O4 = 0,2 * 232 = 46,4g