PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=2,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=280\left(l\right)\)

nC2H4 = 1.12/22.4 = 0.05 (mol) 

CH2 = CH2 + Br2 => CH2Br - CH2Br 

0.05.......................................0.05

mC2H4Br2 = 0.05*188 = 9.4 (g) 

C2H4 + 3O2 -to-> 2CO2 + 2H2O 

0.05.........0.15

Vkk = 5VO2 = 5*0.15*22.4 = 16.8 (l) 

Cho hỏi ạ Br dư thì mình cần bằng la: 

C2H2 + 2Br2--->C2H2Br4

0.05---------------------0.05

mC2H2Br4=0.05×346=17. 3 g

Đúng k 

PTHH: 2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

\(\dfrac{V_{O_2}}{V_{C_2H_2}}=\dfrac{n_{O_2}}{n_{C_2H_2}}=\dfrac{5}{2}\Rightarrow V_{O_2}=2,5.\dfrac{5}{2}=6,25\left(l\right)\)

=> \(V_{kk}=6,25.5=31,25\left(l\right)\)

\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ PTHH:C_2H_4+3O_2-^{t^o}>2CO_2+2H_2O\)

tỉ lệ            1      :    3      :         2      :       2

n(mol)       0,2<-----0,6-------->0,4--------->0,4

\(m_{C_2H_4}=n\cdot M=0,2\cdot28=5,6\left(g\right)\)

Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1\left(mol\right)\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)

Đáp án: A

n _C3H8=0,1 mol

C₃H₈+5O₂-to>3CO₂+4H₂O

0,1-----0,5------------0,3-----0,4 mol

=>V_CO2=0,3.22,4=6,72 l

=>V_O2=0,5.22,4=11,2l

nFe = 16,8/56 = 0,3 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 0,3 ---> 0,2

VO2 = 0,2 . 22,4 = 4,48 (l)

PTHH: 4P + 5O2 -> (t°) 2P2O5

Mol: 0,16 <--- 0,2

mP = 0,16 . 31 = 4,96 (g)