Theo gt ta có: $n_P=0,35(mol)$

$4P+5O_2\rightarrow 2P_2O_5$

a, Ta có: $n_{P_2O_5}=0,175(mol)\Rightarrow m_{P_2O_5}=24,85(g)$

b, Ta có: $n_{O_2}=0,4375(mol)\Rightarrow V_{O_2}=9,8(l)\Rightarrow V_{kk}=49(l)$

c, $2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$

Suy ra $n_{KMnO_4}=0,875(mol)\Rightarrow m_{KMnO_4}=138,25(g)$

a) \(n_{Fe}=\dfrac{3,36}{56}=0,06\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

          0,06->0,04------->0,02

=> m_Fe3O4 = 0,02.232 = 4,64 (g)

b) V_O2 = 0,04.22,4 = 0,896 (l)

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.b.

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

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

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

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

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

\(V_{kk}=4,48.5=22,4l\)

c.

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

      0,4                                                     0,2  ( mol )

\(m_{KMnO_4}=\dfrac{0,4.158}{\left(100-20\right)\%}=79g\)

a/ Số mol Fe là : nFe = 16,8: 56 = 0,3 mol

PTPƯ:

  3Fe   +   2O2 Fe3O4  (1)

0,3 mol → 0,2mol → 0,1 mol

Từ (1) ta có số mol Fe3O4 = 0,1mol

→ m Fe3O4 = n.M = 0,1.232 = 23,2gam

b/ Từ (1) ta có số mol O2 đã dùng nO2 = 0,2 mol

Thể tích khí oxi đã dùng ở đktc: VO2 = n.22,4 = 0,2.22,4 = 4,48 lít

Thể tích không khí đã dùng: Vkk = 5. VO2= 5.4,48 = 22,4 lít.

c/ PTPƯ

2 KMnO4 K2MnO4 + MnO2 + O2 (2)

0,4444mol  ←  0,222mol

Vì lượng Oxi thu được hao hụt 10% nên số mol O2 cần có là:

nO2 = 0,2mol.100/90 = 0.222 mol

Từ (2) ta có số mol KMnO4 = 0,444mol

Khối lượng KMnO4 bị nhiệt phân

mKMnO4 = n.M = 0,444.158 = 70.152 gam

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

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

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 : 

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

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

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

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

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

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

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

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

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

nFe = 5,04 / 56 = 0,09 ( mol)

3Fe + 2O2 --(t^o)-- > Fe3O4

0,09    0,06                   0,03   (mol)

=> mFe3O4 = 0,03 . 232 = 6,9(g)

=> VO2 = 0,06 . 22,4  = 1,344 (l)

=> Vkk = 1,344 . 5 = 6,72(l)

a) 3Fe  +  2O₂      Fe₃O₄  

b) nFe = \(\dfrac{8,4}{56}\)= 0,15 mol 

nFe₃O₄ = \(\dfrac{11,6}{232}\) = 0,05 mol

Ta thấy \(\dfrac{nFe}{3}\)= \(\dfrac{nFe_3O_4}{1}\)=> Fe phản ứng hết 

<=> nO₂ cần dùng = \(\dfrac{2nFe}{3}\)= 0,1 mol 

<=> mO₂ cần dùng = 0,1.32 = 3,2 gam

c) Oxi chiếm thể tích bằng 1/5 thể tích không khí.

Mà V O₂ = 0,1.22,4 = 2,24 lít => V không khí = 2,24 . 5 = 11,2 lít