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Определите массу бутадиена-1, 3, который может быть получен по с.в.лебедева из 150 г 90%-него этилового спиртаподробно)
2 моль 1моль
1)m(C2H5OH) = m(р-ра С2H5OH) * w(C2H5OH)=150 *0,9 =135 г
2) n(C2H5OH) = m / M = 135 / 46г/моль =2,93 моль
3) n(C4H6) = 2,93 / 2 =1,47 моль
4) m(C4H6) = n*M = 1,47 * 54г/моль =79,38 г