maseni udeo

Submitted by danijela on Thu, 05/22/2014 - 15:33
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koliki je maseni udeo slobodne oleinske kiseline ako je neutralizacijom 5g masti potrebno 5cm3 KOH koncentracije 0,025mol/dm3? R:0,705 hvala

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milicaradenkovic - Thu, 05/22/2014 - 17:03

CH3OOC(CH2)7CH=CH(CH2)7CH3CHOOC(CH2)7CH=CH(CH2)7CH3CH2OOC(CH2)7CH=CH(CH2)7CH3+3KOH-------------------->3CH3(CH2)7CH=CH(CH2)7COOK+CH2OHCHOHCH2OH

CH3(CH2)7CH=CH(CH2)7COOK+HCl------------------->KCl+CH3(CH2)7CH=CH(CH2)7COOH

n(KOH)=c*V=0.005*0.025=0.000125 mol

n(kalijumoleata)=n(KOH)=0.000125 mol

n(oleinske kiseline)=n(kalijumoleata)=0.000125 mol

m(oleinske kiseline)=n(oleinske kiseline)*Mr=0.000125*282=0.03525 g

w=m(oleinkse kiseline)/m=0.03525/5=0.00705=0.705 %