Use Cu = 63.5 and molar volume 22.4 L/mol at STP. Copper reacts with dilute nitric acid as 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O. The volume of NO at STP obtained from 12.7 g of copper is:

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Use Cu = 63.5 and molar volume $22.4 L/mol$ at STP. Copper reacts with 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 volume of $NO$ at STP obtained from $12.7 g$ of copper is:

n (Cu) = 12.7/63.5 = 0.2

. Ratio

Cu : NO = 3 : 2

, so

n (NO) = 0.2 \times 2/3 = 0.1333

mol. Volume

= 0.1333 \times 22.4 = 2.99 L