a.\(\%Fe=\dfrac{56.3}{56.3+16.4}.100=72,41\%\)

b.\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)

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

          0,04              0,02     ( mol )

\(m_{O_2}=0,04.32=1,28g\)

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

       0,08                                                   0,04  ( mol )

\(m_{KMnO_4}=0,08.158=12,64g\)

PTHH: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,05\left(mol\right)\\n_{CO_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\\V_{CO_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

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

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,2\left(mol\right)\\n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2\cdot160=32\left(g\right)\\V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)

c) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Zn}=n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,6\cdot65=39\left(g\right)\)

a,

nFe = 22,4/56 = 0,4  (mol)

PTHH

Fe₂O₃ + 3H₂ ---t^o----) 2Fe + 3H₂O        (1)

theo phương trình (1) ,ta có:

nFe₂O₃ = 0,4 x 2 / 1 = 0,8 (mol)

mFe₂O₃ = 160 x 0,8 = 128 (g)

b,

theo pt (1)

nH₂ = (0,4 x 3)/2 = 0,6 (mol)

=) VH₂ = 0,6 x 22,4 = 13,44 (L)

c,

PTHH

Zn + H₂SO₄ -------------) ZnSO₄ + H₂                         (2)

Số mol H₂ cần dùng là 0,6 (mol)

Theo PT (2) :

nZn = nH₂    ==) nZn = 0,6 x 65 = 39 (g)

bài 4

3Fe+2O2-to>Fe3O4

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

n Fe3O4=0,1 mol

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

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

b)

2KMnO4-to>K2MnO4+MnO2+O2

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

=>m KMnO4=0,4.158=63,2g

\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02mol\)

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

0,06  0,04               0,02     ( mol )

\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)

\(m_{O_2}=n_{O_2}.M_{O_2}=0,04.32=1,28g\)

\(pthh:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)

Theo pt: \(n_{Fe}=3.0,02=0,06\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)

Theo pt: \(n_{O_2}=2.0,02=0,04\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,04.32=1,28\left(g\right)\)

a. \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)

PTHH : 3Fe + 2O₂ -t^o-> Fe₃O₄

              0,09    0,06        0,03

\(m_{Fe}=0,09.56=5,04\left(g\right)\)

\(V_{O_2}=0,06.22,4=1,344\left(l\right)\)

b. PTHH : 2KCl + 3O₂ -> 2KClO₃

                            0,06           0,04

\(m_{KClO_3}=0,04.122,5=4,9\left(g\right)\)

4

n _Fe3O4=\(\dfrac{6,96}{232}=0,03mol\)

3Fe+2O₂-to>Fe₃O₄

0,09---0,06-----0,03 mol

=>m _Fe=0,09.56=5,04g

=>V_O2=0,06.22,4=1,344l

b) 

2KClO₃-to>2KCl+3O₂

0,04----------------------0,06 mol

=>m _KClO3=0,04.122,5=4,9g

\(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=3.0,03=0,09\left(mol\right);n_{O_2}=0,03.2=0,06\left(mol\right)\\ a,\Rightarrow m_{Fe}=0,09.56=5,04\left(g\right);V_{O_2\left(đktc\right)}=0,06.22,4=1,344\left(l\right)\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\uparrow\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,06}{3}=0,04\left(mol\right)\\ \Rightarrow b,m_{KClO_3}=122,5.0,04=4,9\left(g\right)\)

a, Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)

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

THeo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\Rightarrow V_{O_2}=0,04.22,4=0,896\left(l\right)\)

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

\(n_{KMnO_4}=2n_{O_2}=0,08\left(mol\right)\Rightarrow m_{KMnO_4}=0,08.158=12,64\left(g\right)\)

a) \(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02\left(mol\right)\).

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

Mol :       3    :     2   :     1

Mol                  0,04 ←  0,02

\(\Rightarrow V_{O_2}=n_{O_2}.22,4=\left(0,04\right).\left(22,4\right)=0,896\left(l\right)\).

b) Từ phương trình ở câu a \(\Rightarrow n_{O_2}=0,04\left(mol\right)\).

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

Mol :           2         :        1         :       1     :   1

Mol :        0,08                    ←                     0,04

\(\Rightarrow m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\left(0,08\right).158=12,64\left(g\right)\).

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

a, \(n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)

b, \(n_{H_2}=3n_{Fe_2O_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{Mg}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,3.24=7,2\left(g\right)\)

a)

\(n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = \dfrac{1}{15}(mol)\\ \Rightarrow m_{Fe_3O_4} = \dfrac{1}{15}.232 = 15,467(gam)\)

b)

\(n_{H_2} = \dfrac{4}{3}n_{Fe} = \dfrac{4}{15}(mol)\\ \Rightarrow V_{H_2} =\dfrac{4}{15}.22,4 = 5,973(lít)\)

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

...............1...............4..............3.....................

...............1/15.........4/15.........0,2..................

a. \(m_{Fe_3O_4}=n_{Fe_3O_4}\cdot M_{Fe_3O_4}=\dfrac{1}{15}\cdot232=\dfrac{232}{15}\left(g\right)\)

b. \(V_{H_2\left(ĐKTC\right)}=n_{H_2}\cdot22,4=\dfrac{4}{15}\cdot22,4=\dfrac{448}{75}\left(l\right)\)