6.35 g of Cu reacts with excess AgNO3 solution: Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag. If 18.0 g of silver is collected, the percent yield is (Cu=63.5, Ag=108)

JEE Main levelSingle correct optionDifficulty 2 of 3mc061

$6.35 g$ of $Cu$ reacts with excess $AgNO X_{3}$ solution: $Cu + 2 AgNO X_{3} Cu (NO X_{3}) X_{2} + 2 Ag$ . If $18.0 g$ of silver is collected, the percent yield is ( $Cu = 63.5$ , $Ag = 108$ )

Cu

= 0.1 mol gives 2 x 0.1 = 0.2 mol

Ag

= 21.6 g (theoretical). Percent yield = 18.0/21.6 x 100 = 83.3%.