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Submitted by tresnja123 on Thu, 12/19/2013 - 10:08
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koliko je potrebno grama joda (Lugolov rastvor) da bi 0,575g apsolutnog etanola u prisustvu alkalnog hidroksida,potpuno izgorelo u jodoformu ?

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milicaradenkovic - Thu, 12/19/2013 - 10:55

CH3CH2OH+6NaOH+4I2---------------->CHI3+5NaI+5H2O+HCOONa

n(CH3CH2OH)=m/Mr=0.575/46=0.0125 mol

n(I2):n(CH3CH2OH)=5:1

n(I2)=5*n(CH3CH2OH)/1=5*0.0125/1=0.0625 mol

m(I2)=n(I2)*Mr=0.0625*254=15.875 g

tresnja123's picture

tresnja123 - Thu, 12/19/2013 - 12:18

a zar ne ide : 

n(I2) : n( C2H5oh) = 4: 1 

? 

zasto 5:1 

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milicaradenkovic - Thu, 12/19/2013 - 15:22

Jeste u pravu si ide 4:1 a ne 5:1 greska u kucanju

n(I2)=4*n(CH3CH2OH)=4*0.0125=0.05 mol

m(I2)=n(I2)*Mr=0.05*254=12.7 g