Question

10 g grasime se trateaza c . 200 g sol de iod de 10%. Excesul de iod se poate aditiona la 0,224 L etena(c.n.). Indicele de iod al grasimii este:
A) 87,3;
B)174,6;
C)127;
D)146;
E)197,3.

Answer 1

CH2=CH2   +   I 2-----> CH2I-CH2I
1mol               1mol
0,224/22,4......0,01mol=----->m=0,01molx(2x127)g/mol=2,57giod se consuma in reactia cu etena
10/100= md/200----> md=20g iod dizolvat in solutie
-masa de iod ce reactioneaza cu 10g grasime: m=20-2,57=>17,43g iod
10g grasime..........17,43g iod
100g.......................indice de iod=174,3;verifica calculele,ia masa iod126,9.....