Bài 27: Điều chế khí oxi-Phản ứng phân hủy

a) \(n_P=\dfrac{46,5}{31}=1,5\left(mol\right)\)

\(4P+5O_2\rightarrow2P_2O_5\)

1,5.....1,875 (mol) 

\(\rightarrow m_{O_2}=1,875.32=60\left(g\right)\)

c. \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(2H_2+O_2\rightarrow2H_2O\)

0,15.....0,075 (mol) 

\(\rightarrow m_{O_2}=0,075.32=2,4\left(g\right)\)

a, \(n_P=\dfrac{m}{M}=1,5\left(mol\right)\)

\(PTHH:4P+5O_2\rightarrow2P_2O_5\)

...............1,5......1,875......

=> \(m_{O_2}=n.M=60\left(g\right)\)

b, \(n_{Al}=\dfrac{m}{M}=2,5\left(mol\right)\)

\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)

................2,5.....1,875.......

=> \(m_{O_2}=n.M=60\left(g\right)\)

c, ( Chắc là ở ĐKTC )

\(n_{H_2}=\dfrac{V}{22,4}=1,5\left(mol\right)\)

\(PTHH:2H_2+O_2\rightarrow2H_2O\)

...............1,5....0,75...........

=> \(m_{O_2}=n.M=24\left(g\right)\)

\(a.\)

\(n_{KClO_3}=\dfrac{3.675}{122.5}=0.03\left(mol\right)\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(0.03........................0.045\)

\(V_{O_2}=0.045\cdot22.4=1.008\left(l\right)\)

\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(\Rightarrow n_{KClO_3}=\dfrac{0.5\cdot2}{3}=\dfrac{1}{3}mol\)

\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}\cdot122.5=40.83\left(g\right)\)

a) Que diêm bùng cháy rất mạnh

b) Que diêm tắt từ từ

a) 

2KMnO4 -to-> K2MnO4 + MnO2 + O2 

0.5__________________________0.25 

VO2 = 0.25*22.4 = 5.6 (l) 

2KClO3 -to-> 2KCl + 3O2 

0.5_______________0.75 

VO2 = 0.75*22.4 = 16.8 (l) 

2KNO3 -to-> 2KNO2 + O2 

0.5________________0.25 

VO2 = 0.25*22.4 = 5.6 (l) 

2HgO -to-> 2Hg + O2

0.5____________0.25 

VO2 = 0.25*22.4 = 5.6 (l) 

b) 

nKNO3 = 50/101 (mol) 

2KNO3 -to-> 2KNO2 + O2 

50/101______________25/101 

VO2 = 25/101 * 22.4 = 5.54 (l) 

nHgO = 50/217 (mol) 

2HgO -to-> 2Hg + O2 

50/217 _________25/217 

VO2 = 2.58 (l) 

rồi sao nữa

- Chắc là viết phương trình :vvv

\(2KClO_3\rightarrow2KCl+3O_2\)

\(a) 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ b) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ c) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ d) C_4H_{10} + \dfrac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O\\ e) 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ g) 2KClO_3 \xrightarrow{^o} 2KCl + 3O_2\)

a) \(2Mg+O_2-^{t^o}\rightarrow2MgO\)

b)\(4P+5O_2-^{t^o}\rightarrow2P_2O_5\)

c)\(2KMnO_4-^{t^o}\rightarrow K_2MnO_4+MnO_2+O_2\)

\(2C_4H_{10}+13O_2\rightarrow8CO_2+10H_2O\)

\(3Fe+2O_2\rightarrow Fe_3O_4\)

\(2KClO_3\rightarrow2KCl+3O_2\)

( Làm nốt :vvv )

\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ \left(mol\right)..0,2\rightarrow.......0,2.......0,3\\ V_{O_2}=0,3.22,4=6,72\left(l\right)\)

\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \left(mol\right)......0,2..\rightarrow.....0,1.........0,1........0,1\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\)

PTHH:2KClO3to→2KCl+3O₂

           .0,2→.......0,2.......0,3 mol

VO2=0,3.22,4=6,72(l)

PTHH:2KMnO₄→K2MnO4+MnO2+O2

......0,2..→.....0,1.........0,1........0,1   mol

VO2=0,1.22,4=2,24(l)

\(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)

a)

\(n_{O_2} = \dfrac{48}{32} = 1,5(mol)\)

Theo PTHH : 

\(n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 1(mol)\\ \Rightarrow m_{KClO_3} = 1.122,5 = 122,5(gam)\)

b)

\(n_{O_2} = \dfrac{44,8}{22,4} = 2(mol) \)

Theo PTHH : 

\(n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{4}{3}mol\\ \Rightarrow m_{KClO_3} = \dfrac{4}{3}.122.5 = 163,33(gam)\)

nO2=44,822,4=2(mo

a)  PTHH: 2KClO3-->2KCl + 3O2                            1mol <---1mol<---1,5mol  nO2=48/32=1,5mol=> nKClO3=1 =>mKClO3=1.122,5=122,5 gamvậy cần 122,5 gam KClO3 để điều chế 48 gam khí oxi

b)PTHH: 2KClO3-->2KCl + 3O2                   4/3mol <---4/3mol<---2mol  nO2=44,8/22,4=2mol=> nKClO3=4/3mol =>mKClO3=4/3.122,5=163,33333gamVậy cần 163,33333gam KClO3 để điều chế 44,8 lít khí oxi ở đktc

\(n_P = \dfrac{3,1}{31} = 0,1(mol)\)

\(4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)

Theo PTHH : 

\(n_{O_2\ pư} = \dfrac{5}{4}n_P =0,125(mol)\\ n_{P_2O_5} = 0,5n_P = 0,05(mol)\)

Vậy :

\(V_{O_2}= 0,125.22,4 = 2,8(lít)\\ m_{P_2O_5} = 0,05.142 = 7,1(gam)\)

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

Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,3}{5}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,125\left(mol\right)\\n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow V_{O_2\left(pư\right)}=0,125.22,4=2,8\left(l\right)\)

\(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

Bạn tham khảo nhé!