C=md/ms*100=>md=9,6g CH3-COOH
n=m/M=9,6/60=0,16 moli acid
a. H3C-COOH + Na => H3C-COONa + 1/2 H2
b. n acid=nNa=0,16 moli Na
m Na = 0,16*23=3,68g pur
p=mp/mimp*100=>mimp=4,6g Na impur
c. n acid = n acetat
Macetat = 82g/mol
m acetat = 82*0,16=13,12g (mt)
n(eta)=mp/mt*100=>mp=9,84g acetat