Alkohol

Submitted by Immortalius on Tue, 06/24/2014 - 20:16
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Koliko se grama vode izdvoji iz potrebne kolicine odgovarajuceg hidroksilnog derivata ugljovodonika pri dobijanju 0.42 propena?

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milicaradenkovic - Wed, 06/25/2014 - 08:03

CH3CH2CH2OH----------------------->CH3CH=CH2+H2O

n(propena)=m/Mr=0.42/42=0.01 mol

n(vode)=n(propen)=0.01 mol

m(vode)=n(vode)*Mr=0.01/18=0.18 g