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ilya1046
@ilya1046
August 2021
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сколько литров углекислого газа выделится при сгорание 4 моль пропилена
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Chemistry2016
2CH₃-CH=CH₂ + 9O₂ → 6CO₂ + 6H₂O
n(CH₃-CH=CH₂):n(CO₂) = 2:6
n(CO₂) = 4x3 = 12 моль
V(CO₂) = 12x22,4 = 268,8 л
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Answers & Comments
n(CH₃-CH=CH₂):n(CO₂) = 2:6
n(CO₂) = 4x3 = 12 моль
V(CO₂) = 12x22,4 = 268,8 л