Bài 1:

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

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

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{3}< \dfrac{0,1}{2}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,1-\dfrac{1}{15}=\dfrac{1}{30}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\dfrac{1}{30}.32\approx1,067\left(g\right)\\V_{O_2\left(dư\right)}=\dfrac{1}{30}.2,24\approx0,746\left(l\right)\end{matrix}\right.\)

b, Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)

Bài 2:

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

a, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232\approx15,467\left(g\right)\)

b, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{15}\left(mol\right)\)

\(\Rightarrow V_{O_2}=\dfrac{2}{15}.22,4\approx2,9867\left(l\right)\)

c, PT: \(2N_2+5O_2\underrightarrow{t^o}2N_2O_5\)

Ta có: \(n_{N_2}=\dfrac{2,8}{28}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{\dfrac{2}{15}}{5}\), ta được N₂ dư.

Theo PT: \(n_{N_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{4}{75}\left(mol\right)\)

\(\Rightarrow m_{N_2O_5}=\dfrac{4}{75}.108=5,76\left(g\right)\)

Bạn tham khảo nhé!

Bài 1 : 

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(n_{O_2}=\dfrac{2.24}{224}=0.1\left(mol\right)\)

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

\(Bđ:0.1......0.1\)

\(Pư:0.1.......\dfrac{1}{15}...\dfrac{1}{30}\)

\(Kt:0........\dfrac{1}{30}....\dfrac{1}{30}\)

\(V_{O_2\left(dư\right)}=\dfrac{1}{30}\cdot22.4=0.747\left(l\right)\)

\(m_{Fe_3O_4}=\dfrac{1}{30}\cdot232=7.73\left(g\right)\)

Bài 2 : 

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2.......0.3.......\dfrac{1}{15}\)

\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.47\left(g\right)\)

\(n_{N_2}=\dfrac{2.8}{28}=0.1\left(mol\right)\)

\(2N_2+5O_2\underrightarrow{t^0}2N_2O_5\)

\(0.12......0.3........0.12\)

\(m_{N_2O_5}=0.12\cdot108=12.96\left(g\right)\)

Bài 3:

Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

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

_____0,2____0,6____0,4 (mol)

\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(m_{Fe}=0,4.56=22,4\left(g\right)\)

Bài 4:

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

\(n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,35}{5}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,35-0,25=0,1\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,1.32=3,2\left(g\right)\)

b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

Lần sau bạn nên chia nhỏ câu hỏi ra nhé.

Bài 1:

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

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

Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

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

\(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KMnO_4}=0,4.158=63,2\left(g\right)\)

Bài 2:

a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

_____0,1___________0,1_____0,15 (mol)

\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PT: \(n_{CuO}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)

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

a, Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,075\left(mol\right)\)

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

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

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

Bạn tham khảo nhé!

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

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

gọi số mol Fe là x 
pthh : 3Fe + 2O2 --t---> Fe3O4 
          x---->\(\dfrac{2}{3}\)x----------> \(\dfrac{x}{3}\) 
=> mFe3O4= \(\dfrac{x}{3}\) . 232 = \(\dfrac{232x}{3}\) (G)
=> VO2 = \(\dfrac{2x}{3}\) . 22,4 = \(\dfrac{224x}{15}\) (L)

Số liệu ??

không có đâu bạn m như là x ấy 

Bạn tách ra từng câu nhé!

Bài 3.

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{36}{56}=0,6428mol\)

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

0,6428 ----- 0,4285           ( mol )

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

0,857                                                      0,4285    ( mol )

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,857.158=135,406g\)

Bài 4.

a.\(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{51}{102}=0,5mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

  1       0,75            0,5     ( mol )

\(m_{Al}=n_{Al}.M_{Al}=1.27=27g\)

\(V_{O_2}=n_{O_2}.22,4=0,75.22,4=16,8l\)

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

       1,5                                                      0,75   ( mol )

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=1,5.158=237g\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

  0,5                                0,75   ( mol )

\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,5.122,5=61,25g\)

Bài 5.

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{14,4}{24}=0,6mol\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

0,6 <   0,2                         ( mol )

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

\(m_{MgO}=n_{MgO}.M_{MgO}=0,2.40=8g\)

Bài 6.

\(n_{Cu}=\dfrac{m_{Cu}}{M_{Cu}}=\dfrac{12,8}{64}=0,2mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

 0,2     0,1            0,2  ( mol )

\(V_{kk}=V_{O_2}.5=\left(0,1.22,4\right).5=2,24.5=11,2l\)

\(m_{CuO}=n_{CuO}.M_{CuO}=0,2.80=16g\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\) 
            0,15     0,1       0,05 
\(m_{Fe_2O_4}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.11,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\) 
                0,2                                                0,1 
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\) 
            0,15    0,1      0,05 
\(m_{Fe_3O_{\text{ 4}}}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) 
              0,1                                               0,05 
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)

đề bảo đốt trong khí oxi nhưng bắt tính V_kk :) ?

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

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

0,2      2/15     1/15            ( mol )

\(V_{kk}=V_{O_2}.5=\left(\dfrac{2}{15}.22,4\right).5=14,93l\)

\(m_{Fe_3O_4}=\dfrac{1}{15}.232=15,46g\)

\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\)

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

 0,03   0,02               0,01  ( mol )

\(m_{Fe_3O_4}=0,01.232=2,32\left(g\right)\)

\(V_{kk}=0,02.22,4.5=2,24\left(l\right)\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

   \(\dfrac{1}{75}\)                               0,02   ( mol )

\(m_{KClO_3}=\dfrac{1}{75}.122,5=1,63\left(g\right)\)

a,nFe=1,68/56=0,03 mol

Ta có PTHH : 3Fe + 2O2 --> Fe3O4 (1) ( ở trên dấu --> có to nha )

Theo PTHH ta có :

nFe3O4=1/3nFe=1/3.0,03=0,01 mol

nO2=2/3nFe=2/3.0,03=0,02 mol

=>mFe3O4= 0,01.232=2,32g

=>Vkk=5.(0,02.22,4)=2,24 l

b, Ta có PTHH: 2KClO3 --> 2KCl + 3O2 (2) ( trên dấu --> vẫn có to )

Gọi x là số mol KClO3 cần dùng ( x > 0 )

Theo PTHH (3) và theo bài ra ta có PTHH sau:

2/3x=0,02 

=> x=0,03 mol

=> mKClO3= 0,03.122,5= 3,675g