Proszę o pomoc zad 3 4 i 5
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a)C9H20 + 14O₂ --> 9CO₂ + 10H2O
b) 2C8H14 + 15O₂ --> 16CO + 14H2O
c) 2C7H14 + 7O₂ ---> 14C + 14H2O
zad.2
HC≡C-CH₃(propyn) +2H₂ --> H₃C - CH₂ - CH₃ (propan)
zad3
MC₈H₁₆ = (8*12u) + 16*1u= 96u+ 16u =112u
proporcja:
112u-----100%
96u ---x(%C)
x= 100%*96u/ 112u
x=85,7%