Zinc reacts with dilute nitric acid to give $Zn (NO X_{3}) X_{2}$ , $N X_{2} O$ and water. After balancing the equation by the electron-change method, the percentage of the nitric acid consumed that acts as the oxidising agent (i.e. is reduced) is

Answer: D

Zn Zn X^{2 +}

gives 2 e

^{-}

;

2 N X^{+ 5} N X_{2} X^{+ 1} O

takes 8 e

^{-}

. So

4 Zn

per

N X_{2} O

4 Zn + 10 HNO X_{3} 4 Zn (NO X_{3}) X_{2} + N X_{2} O + 5 H X_{2} O

. Of 10

HNO X_{3}

, 2 are reduced to

N X_{2} O

and 8 stay as nitrate, so the fraction reduced =

2/10 = 20%