6.35 g of copper reacts with excess concentrated nitric acid: Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O. All the NO2 is then passed into water, where 3NO2 + H2O -> 2HNO3 + NO. The volume of NO at STP is (Cu=63.5)

JEE Advanced levelSingle correct optionDifficulty 3 of 3mc152

6.35 g of copper reacts with excess concentrated nitric acid: $Cu + 4 HNO X_{3} Cu (NO X_{3}) X_{2} + 2 NO X_{2} + 2 H X_{2} O$ . All the $NO X_{2}$ is then passed into water, where $3 NO X_{2} + H X_{2} O 2 HNO X_{3} + NO$ . The volume of $NO$ at STP is ( $Cu = 63.5$ )

Moles of Cu = 6.35/63.5 = 0.1. First step gives 2 x 0.1 = 0.2 mol

NO X_{2}

. Second step: 3 mol

NO X_{2}

gives 1 mol NO, so NO = 0.2/3 = 0.0667 mol. Volume = 0.0667 x 22.4 = 1.49 L.