Question

a .5674g piece of copper is added to 10.00mL of 16M HNO3, producing ..8024g copper(II) nitrate. WHat is the percent yield of the reaction?

Answer 1

The reaction is   Cu (s) + 2HNO3(aq) ---> Cu(NO3)2 + H2

Cu moles = mass of Cu / Molar mass of Cu = 0.5674 / 63.546 = 0.008929

HNO3 moles = M x V = 16 x ( 10/1000) =0.16   ( where V = 10 ml = 10/1000 L )

as per reaction 1Cu reacts with 2HNO3 , hence HNO3 moles required per 0.008929 moles Cu is 0.008929 x 2 = 0.01786 but we have 0.16 moles , which means HNO3 is in excess.

Hence Cu is limiting reagnet since present in lesser relatively than HNO3

Product depends on limiting reagent moles

Moles of Cu(NO3)2 = Moles of Cu reacted = 0.008929

CuNO3)2 mas s= moels x moalr mass of copper (II) nitrate = 0.008929 x 187.56 = 3.35 g

this is our theoretical yiled , but obtained yiled = 0.8024 g

percentyiled = ( 100 x obtained yield/ theoretical yiled) = ( 100 x 0.8024 /3.35) = 23.96 %