Pomoc oko zadatka I

Submitted by MiaHemija on Sat, 12/07/2013 - 09:26
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Opšta diskusija

1.Odredi masu 3%-nog rastvora natrijum-hidroksida koji se utrosi ya potpunu neutraliyaciju 50g 5%-nog sirceta.

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milicaradenkovic - Sat, 12/07/2013 - 09:54

CH3COOH+NaOH------------------->CH3COONa+H2O

ms=w*mr=50*0.05=2.5 g CH3COOH

n(CH3COOH)=ms/Mr=2.5/60=0.0417 mol

n(NaOH)=n(CH3COOH)=0.0417 mol

m(NaOH)=n(NaOH)*Mr=0.0417*40=1.668 g

mr=m(NaOH)/w=1.668/0.03=55.6 g 3% rastvora