a)

\(n_{O_2} = \dfrac{5,6}{22,4} = 0,25(mol)\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)

Theo PTHH :

\(n_{Fe} = \dfrac{3}{2}n_{O_2} = 0,375(mol)\\\Rightarrow a = 0,375.56 = 21(gam)\)

b)

Theo PTHH :

\(n_{Fe_3O_4} = 0,5n_{O_2} = 0,25.0,5 = 0,125(mol)\\\Rightarrow m_{Fe_3O_4} = 0,125.232 = 29(gam)\)

Bài 3 : 

PTHH :  \(6Fe+4O_2\left(t^o\right)->2Fe_3O_4\)      (1)

\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{2,32}{56.3+16.4}=0,01\left(mol\right)\)

Từ (1) => \(3n_{Fe_3O_4}=n_{Fe}=0,03\left(mol\right)\)

=> \(m_{Fe}=n.M=1,68\left(g\right)\)

Từ (1) => \(2n_{Fe_3O_4}=n_{O_2}=0,02\left(mol\right)\)

=> \(V_{O_2\left(đktc\right)}=n.22,4=0,448\left(l\right)\)

Bài 4 : 

PTHH : \(4P+5O_2\left(t^o\right)->2P_2O_5\)    (1)

\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{32}=0,21\left(mol\right)\)

Có : \(n_P< n_{O_2}\left(0,2< 0,21\right)\)

-> P hết ; O₂ dư

Từ (1) -> \(\dfrac{1}{2}n_P=n_{P_2O_5}=0,1\left(mol\right)\)

=> \(m_{P_2O_5}=n.M=14,2\left(g\right)\)

Bài 3:

\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

PTHH: 3Fe + 2O₂ ---t^o→ Fe₃O₄

Mol:    0,03     0,02            0,01

\(m_{Fe}=0,03.56=1,68\left(g\right);V_{O_2}=0,02.22,4=0,448\left(l\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

           0,6<--0,4

=> m_Fe = 0,6.56 = 33,6(g)

\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.6......0.4\)

\(m_{Fe}=0.6\cdot56=33.6\left(g\right)\)

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(Fe_3O_4+4CO\underrightarrow{^{t^0}}3Fe+4CO_2\)

\(..................0.3....0.4\)

\(m_{CO_2}=0.4\cdot44=17.6\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH :Fe₂O₃ + 3H₂ --t^o-> 2Fe + 3H₂O

              0,15      0,45      0,3

\(V_{O_2}=0,45.22,4=10,08\left(l\right)\)

\(m_{Fe}=0,3.56=16,8\left(g\right)\)

Chọn D

\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

              0,15-->0,45------>0,3

V = 0,45.22,4 = 10,08 (l)

m = 0,3.56 = 16,8 (g)

=> D

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

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

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

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

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

c. \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

0,3 >   0,15                        ( mol )

0,225   0,15             0,075  ( mol )

\(m_{Fe_3O_4}=0,075.232=17,4\left(g\right)\)

d. \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

\(Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)

0,075  <   0,35                                ( mol )

0,075        0,3                             ( mol )

Chất dư là H₂

\(m_{H_2\left(dư\right)}=\left(0,35-0,3\right).2=0,1\left(g\right)\)   

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

\(0.3......0.2.........0.1\)

\(V_{O_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{Fe_3O_4}=0.1\cdot232=23.2\left(g\right)\)

A

A

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 \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)