Alkoholi

Submitted by Vuk Biljuric on Sat, 06/28/2014 - 11:38
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Koliko ima primarnih alkohola cijih 3,7g u reakciji sa natrijumom oslobadja 560cm3 gasa???

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milicaradenkovic - Sat, 06/28/2014 - 21:22

2CnH(2*n+1)OH+2Na--------------------->2CnH(2*n+1)ONa+H2

n(H2)=V(H2)/Vm=0.560/22.4=0.025 mol

n(akohola)=2*n(H2)=2*0.025=0.05 mol

Mr=m/n(alkohola)

Mr=3.7/0.05=74 g/mol

Mr=n*Ar(C)+(2*n+1)*Ar(H)+Ar(H)+Ar(O)

Mr=12*n+2+2*n+16

74=14*n+18

14*n=56

n=56/14=4 i formula je C4H10O

1. CH3CH2CH2CH2OH                 n-butanol

2. CH3CH(CH3)CH2OH                 2-metil- propanol