Question

Prin bromurarea avansata a 500g fenol de puritate 94% se ibtine un precipitat alb 2,4,6 tribromofenol. Stiind ca reactia are loc cu un randament de 80% sa se calculeze masa de precipitat obtinuta si masa solutiei de apa de brom c=8% necesara procesului. Reactia o stiu..dar rezolvarea..

Answer 1

p=mp/mimp*100=>mp=0,94*500=470g C6H5-OH pur

n=m/M=470/94=5 moli fenol

C6H5-OH + 3Br2 => 2,4,6-tribromofenol (C6H2(OH)Br3] + 3HBr

teoretic : nC6H2Br3-OH=nC6H5-OH = 5 moli

m C6H2Br3-OH = 5*331 = 1655g

practic: mp=80*1655/100=1324g

nBr2=3nC6H5-OH = 3*5 = 15 moli

md Br2 = 15*160=2400g

ms = 240000/8 = 30000g = 30kg sol Br2