Use Cu = 63.5, Ag = 108 and molar volume 22.4 L/mol at STP. A 5.00 g Cu-Ag alloy contains 40.0% Cu by mass. It dissolves completely in dilute nitric acid, 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O and 3Ag + 4HNO3 -> 3AgNO3 + NO + 2H2O. Find the volume of NO evolved at STP in mL (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3rx156

Use Cu = 63.5, Ag = 108 and molar volume $22.4 L/mol$ at STP. A $5.00 g$ Cu-Ag alloy contains $40.0%$ Cu by mass. It dissolves completely in dilute nitric acid, $3 Cu + 8 HNO X_{3} 3 Cu (NO X_{3}) X_{2} + 2 NO + 4 H X_{2} O$ and $3 Ag + 4 HNO X_{3} 3 AgNO X_{3} + NO + 2 H X_{2} O$ . Find the volume of $NO$ evolved at STP in mL (nearest integer).

Numerical answer type, answer in mL, tolerance 1.

= 2.00/63.5 = 0.03150

mol gives

\times 2/3 = 0.02100

mol NO. Ag

= 3.00/108 = 0.02778

mol gives

\times 1/3 = 0.00926

mol NO. Total

= 0.03026

mol. Volume

= 0.03026 \times 22400 = 678 mL