a)\(n_{Fe}=\dfrac{22,4}{56}=0,4mol\)

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

0,4     \(\dfrac{4}{15}\)      \(\dfrac{2}{15}\)

\(V_{O_2}=\dfrac{4}{15}\cdot22,4=5,973l\)

b)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

   \(\dfrac{8}{45}\)                          \(\dfrac{4}{15}\)

  \(m_{KClO_3}=\dfrac{8}{45}\cdot122,5=21,78g\)

a)

PTHH: 3Fe + 2O₂  ^{____\(t^o\)____}> Fe₃O₄  (1) 

b) Ta có: n_Fe  = \(\dfrac{25.2}{56}=0.45\left(mol\right)\)

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) \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}0.3=0.2\left(mol\right)\)

=> \(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{23,2}{232}=0,1mol\)

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

0,3       0,2              0,1          ( mol )

\(m_{Fe}=n_{Fe}.M_{Fe}=0,3.56=16,8g\)

\(V_{O_2}=n_{O_2}.22,4=0,2.22,4=4,48l\)

\(V_{kk}=\dfrac{4,48.100}{20}=22,4l\)

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

    0,4                                         0,2               ( mol )

\(n_{KMnO_4}=\dfrac{0,4}{85\%}=\dfrac{8}{17}mol\)

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\dfrac{8}{17}.158=74,3529g\)

3Fe+2O₂-to>Fe₃O₄

0,3------0,2-------0,1mol

n Fe3O4=\(\dfrac{23,2}{232}\)=0,1 mol

->m Fe=0,3.56=16,8g

b)

VO2=0,2.22,4=4,48l

=>Vkk=4,48.5=22,4l

c)2KMnO₄-to>K₂MnO₄+MnO₂+O₂

   0,4-------------------------------------0,2

=>m KMnO4=0,4.158.\(\dfrac{100}{85}\)=74,35g

\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\) 
           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 
=> \(m_{KClO3}=122,5\left(g\right)\)

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

\(m_{KClO_3}=n.M=0,6.\left(39+35,5+16.3\right)=55,5\left(g\right).\)

3Fe+2O₂-to>Fe₃O₄

0,225--0,15

n Fe=\(\dfrac{12,6}{56}\)=0,225 mol

VO2=0,15.22,4=3,36l

2KClO₃-to>2KCl+3O₂

0,1---------------------0,15

=>m _KClO3=0,1.122,5=12,25g

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

\(b,\)

Ta có : \(n_{Fe}=\dfrac{m}{M}=\dfrac{126}{56}=2,25\left(mol\right)\)

\(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.2,25=1,5\left(mol\right)\)

\(\Rightarrow VO_2=33,6\left(l\right)\)

\(c,\)

\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)

Theo \(PTHH:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.1,5=1\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=n.M=1,122,5=122,5\left(g\right)\)

\(a,CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 Vì n và V tỉ lệ thuận với nhau. Nên ta có:

\(V_{O_2}=2.V_{CH_4}=2.2,768=5,536\left(l\right)\)

\(b,V_{kk}=\dfrac{100}{21}.V_{O_2}=\dfrac{100}{21}.5,536=\dfrac{2768}{105}\left(l\right)\)