aminokiseline

Submitted by Vesna on Sat, 07/05/2014 - 22:09
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22. Koliko ce se dobiti molova alanina  dejstvom 5600 ml amonijaka (normalni uslovi) na odgovarajuci alfa-halogensku kiselinu?

Resenje 0,125

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milicaradenkovic - Sun, 07/06/2014 - 09:52

CH3CHClCOOH+2NH3----------------->CH3CHNH2COOH+NH4Cl

n(NH3)=V(NH3)/Vm=5.6/22.4=0.25 mol

n(alanina)=n(NH3)/2=0.25/2=0.125 mol