Question

If 1.6 mol cu metal reacts with a solution containing 3.9 mol AgNO3, what ions will be present at the end of the reaction?

Cu(s) + 2AgNO3(aq) --> Cu(NO3)2(aq) + Ag(s)

(please be as detailed in your steps as possible)

Answer 1

Cu (s)   +    2 AgNO3 (aq) ----------------------> Cu(NO3)2 (aq) + 2 Ag (s)

1 mole            2 mole                                             1 mole            2 mole

1.6                   3.9                                                  ?

Cu moles 1.6

AgNO3 moles = 3.9 /2 = 1.95

so AgNO3 is the excess reagent and Cu is limiting reagent

so total Cu will be consumed and Cu(NO3)2 formed = 1.6 moles

here now AgNO3 is left = 3.9 - 2 x 1.6 = 0.3

after reaction remaining species AgNO3 , Cu(NO3)2 and Ag

ions present   :   Ag⁺ , Cu²⁺ , NO₃^-