Question

Answer 1

Bài 1 : 

\(n_{C2H4}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

a) Pt : \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O|\)

             1           3           2          2

            0,1       0,3

b) \(n_{O2}=\dfrac{0,1.3}{1}=0,3\left(mol\right)\)

\(V_{O2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) \(V_{KK}=5.V_{O2}=5.6,72=33,6\left(l\right)\)

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Answer 2

C₂H₄+3O₂-to>2CO₂+2H₂O

0,1---------0,3 mol

n _C2H4=\(\dfrac{2,479}{24,79}\)=0,1 mol

=>V_O2=0,3.22,4=6,72l

=>V_kk=6,72.5=33,6l

 

Answer 3

nC2H4 = 2,479/24,79 = 0,1 (mol)

PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O

nO2 = 0,1 . 3 = 0,3 (mol)

VO2 = 0,3 . 24,79 = 7,437 (l)

Vkk = 7,437 : 20% = 37,185 (l)