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

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

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

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

Theo pt: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}\cdot0,1=\dfrac{1}{15}mol\)

\(\Rightarrow V_{O_2}=\dfrac{1}{15}\cdot22,4=1,5l\)

Cũng theo pt: \(n_{Fe_3O_4}=\dfrac{1}{3}\cdot0,1=\dfrac{1}{30}mol\)

\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}\cdot232=7,73g\)

a. \(n_{Fe}=\dfrac{5.6}{56}=0,1\left(mol\right)\)

PTHH : 3Fe + 2O₂ ----t^o----> Fe₃O₄

             0,1       \(\dfrac{0.2}{3}\)                 \(\dfrac{0.1}{3}\)

b. \(V_{O_2}=\dfrac{0.2}{3}.22,4=\dfrac{4.48}{3}\left(l\right)\)

c. \(m_{Fe_3O_4}=\dfrac{0.1}{3}.232=\dfrac{23.2}{3}\left(g\right)\)

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

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

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

nFe = 16,8 : 56 = 0,3  (mol) 
pthh :3 Fe + 2O2 -t--> Fe3O4
           0,3--------------> 0,1 (mol) 
=> mFe3O4 =0,1 . 232 = 23,2(G) 
nH2 = 44,8 : 22,4 = 2 (g) 
pthh : Fe3O4 + H2 -t--> Fe + H2O
   LTL :  0,1 / 1  <  2 /1 
=> H2 du 
nH2 (pu) = nFe3O4 = 0,1 (mol) 
=> nH2 (d) = 2-0,1 = 1,9 (mol) 
mH2 (d) = 1,9 . 2 = 3,8 (g) 

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

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

\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ Mol:0,4\rightarrow\dfrac{4}{15}\rightarrow\dfrac{2}{15}\)

\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\rightarrow V_{kk}=\dfrac{448}{75}.5=\dfrac{448}{15}\left(l\right)\\m_{Fe_3O_4}=\dfrac{2}{15}.232=\dfrac{464}{15}\left(g\right)\end{matrix}\right.\)

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 ---t^o---> 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 ---t^o---> 2KCl + 3O₂

Mol:       \(\dfrac{0,16}{9}\)                            \(\dfrac{0,8}{3}\)

\(m_{KClO_3}=\dfrac{0,16}{9}.122,5=2,178\left(g\right)\)

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
=> pthh :\(3Fe+2O_2-t^o->Fe_3O_{\text{4 }}\)

             0,4       0,26              0,13 
=> \(V_{O_2}=0,26.22,4=5,973\left(L\right)\)
=> \(V_{KK}=5,973:\dfrac{1}{5}=29,86\left(L\right)\)
\(m_{Fe_3O_4}=0,13.232=30,16\left(g\right)\)
\(pthh:2KClO_3-t^o->2KCl+3O_2\)
           0,173                                0,26 
=> \(m_{KClO_3}=0,173.122,5=21,1925\left(G\right)\)

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

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

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

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

b) \(V_{O_2\left(\text{đ}kc\right)}=0,4.24,79=9,916\left(l\right)\)

c) PTHH: \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)

                \(\dfrac{4}{15}\)<-------------------0,4 (mol)

=> \(m_{KClO_3}=\dfrac{4}{15}.122,5=\dfrac{98}{3}\left(g\right)\)

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

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