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YanaSavenkova
@YanaSavenkova
July 2022
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Какой объем этана можно получить при взаимодействии 151,5 г. хлорметана с металлическим натрием?
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Potassium
2CH3Cl + 2Na = 2NaCl + C2H6
151.5/101 = x/22.4
x = (151.5*22.4)/101 = 33.6 л
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Answers & Comments
151.5/101 = x/22.4
x = (151.5*22.4)/101 = 33.6 л