Question

Napisz równania reakcji spalania(po trzy: całkowite i niecałkowite) a) metanu b) etanu c) propanu d) etenu e) etinu f) butinu g) alkoholu etylowego h) kwasu octowego i) gliceryny j) octamu etylu Proszę o pomoc :)

Answer 1

a) CH4 + 2O2 <=> CO2 + 2H2O

    2CH4 + 3O2 <=> 2CO + 4H2O

    CH4 + O2 <=> C + 2H2O

b) 2C2H6 + 7O2 <=> 4CO2 + 6H2O

    2C2H6 + 5O2 <=> 4CO + 6H2O

    2C2H6 + 3O2 <=> 4C + 6H2O

c) C3H8 + 5O2 <=> 3CO2 + 4H2O

    2C3H8 + 7O2 <=> 6CO + 8H2O

    C3H8 + 2O2 <=> 3C + 4H2O

d) C2H4 + 3O2 <=> 2CO2 + 2H2O

C2H4 + 2O2 <=> 2CO + 2H2O

C2H4 + O2 <=> 2C + 2H2O

e) 2C2H2 + 5O2 <=> 4CO2 + 2H2O

2C2H2 + 3O2 <=> 4CO + 2H2O

2C2H2 + O2 <=> 4C + 2H2O

f) 2C4H6 + 11O2 <=> 8CO2 + 6H2O

2C4H6 + 7O2 <=> 8CO + 6H2O

2C4H6 + 3O2 <=> 8C + 6H2O

g)CH3CH2OH + 3O2 <=> 2CO2 + 3H2O

CH3CH2OH + 2O2 <=> 2CO + 3H2O

CH3CH2OH + O2 <=> 2C + 3H2O

h)CH3COOH + 2O2 <=> 2CO2 + 2H2O

CH3COOH + O2 <=> 2CO + 2H2O

CH3COOH + O2 -> C + 2H2O

i) 2C3H8O3 + 7O2 <=> 6CO2 + 8H2O
C3H8O3 + 2O2 <=> 3CO + 4H2O

2C3H8O3 + O2 <=> 6C + 8H2O

j) CH3COOC2H5 + 5O2 <=> 4CO2 + 4H2O

CH3COOC2H5 + 3O2 <=> 4CO + 4H2O

CH3COOC2H5 + O2 <=> 4C + 4H2O