#stackrel(0)Cu+Hstackrel(+5)NO_3->stackrel(+2)Cu(NO_3)_2+stackrel(+2)NO+H_2O#
Change in Oxidation Number
#Cu(0)->Cu(+2)=>"2 unit oxidation"#
#N(+5)->N(+2)=>"3 unit reduction"#
#"Reacting Ratio of " Cu(+2):N(+5)=3:2#
3mol #Cu(NO_3)_2# formed from 3mol Cu requires extra 6 mol #HNO_3# as salt former which is added in LHS making total 8 mol #HNO_3# in reactant side.The RHS is then adjusted as required.
#3stackrel(0)Cu+8Hstackrel(+5)NO_3->3stackrel(+2)Cu(NO_3)_2+2stackrel(+2)NO+4H_2O#
Another Way
(1) First write the balanced equation for decomposition reaction of
#HNO_3# forming the oxide of Nitrogen produced, water and oxygen as follows
#2HNO_3->H_2O+2NO+3O#
(2) Adjust the O-atom produced in step (1) combining with required proportion of Cu atoms as follows
#3Cu+3O->3CuO#
(3) Write balanced equation of reaction of 3CuO with #HNO_3# forming salt and water
#3CuO+6HNO_3->3Cu(NO_3)_2+3H_2O#
(4) Finally add those equations to have the required balance equation.
#2HNO_3->H_2O+2NO+cancel(3O)#
#3Cu+cancel(3O)->cancel(3CuO)#
#cancel(3CuO)+6HNO_3->3Cu(NO_3)_2+3H_2O#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#3Cu+8HNO_3->3Cu(NO_3)_2+2NO+4H_2O#
Shortcut tips for balancing the equations of reactions between metal and nitric acids in various cases where oxides of nitrogen are produced.
(1) Let moles #Cu# be #x# and moles of #HNO_3# be #y# in the balanced equation of a reaction.
Then #x=5-"ON of N in its oxide formed"#
As for above reaction #x=5-2=3#
(2) Then #y=x xxv+v#,
#"where" v = "valency of metal or OS of metal in its nitrate salt "#
So for above reaction #v=2#
and #y=3xx2+2=8#
(3) Knowing the value of x and y we can easily balance the equation.
