Question

How many grams of urea on heating yield 10²² molecules of biuret by the reaction :
2 CO (NH2)2 -
>HN - CO - NH - CO - NH, + NH, _ BTU 9e.
(A) 1.495
(B) 0.995
(C) 1.99
(D) 1.​

Answer 1

Answer:

2 CO (NH2)2⟶ H2N - CO - NH - CO - NH2 + NH3 ? ... They will be obtained by heating 2×0.0166=0.033 moles of urea.