Question

O cantitate de 80,75 g cauciuc butadienacrilonitrilic decoloreaza 500g solutie de brom de concentratie 40%. Raportul molar dintre monomeri (butadiena:acrilonitril) este?

Answer 1

-(CH2-CH=CH-CH2-)x--(CH(CN)-CH2--)y
-calculez masa si nr. moli de brom aditionat de componenta care are 1dubla legatura( 1 dubla legatura aditioneaza 1mol brom)
c/100=md/ms
40/100= md/500---> md= 200g
M,Br2=160g/mol--> n=m/M; n= 200g/160g/mol=1,25molBr2--> x=1,25
-M,monomerul butadienic= 54g/mol
1mol brom........54g
1,25mol...........m= 67,5g componenta butadienica
-componenta acrilonitrilica are masa  m= 80,75-67,5= 13,25g
si M= 53g/mol---> n= y=13,25g/53g/mol=0,25
-raportul cerut este  x/y= 1,25/0,25=5/1