Copper reacts with cold dilute nitric acid as 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O. The ratio of moles of HNO3 acting as an acid to moles acting as an oxidising agent is

JEE Main levelSingle correct optionDifficulty 2 of 3pn170

Copper reacts with cold dilute nitric acid as $3 Cu + 8 HNO X_{3} 3 Cu (NO X_{3}) X_{2} + 2 NO + 4 H X_{2} O$ . The ratio of moles of $HNO X_{3}$ acting as an acid to moles acting as an oxidising agent is

Of 8 mol

HNO X_{3}

, 6 mol appear as

NO X_{3} X^{-}

Cu (NO X_{3}) X_{2}

(acid role) and 2 mol are reduced to

NO

(oxidant role). The ratio is

6 : 2 = 3 : 1