If the questions is Cu2+ ions in this experiment are produced by the reaction of 1.0 g copper turnings with excess nitric acid and I am asked how many moles of Cu2+ are produced. I first have to write the equation right? Is this the right equation? Cu(s)+HNO3(aq)- Cu2+

Question

If the questions is Cu2+ ions in this experiment are produced by the reaction of 1.0 g copper turnings with excess nitric acid and I am asked how many moles of Cu2+ are produced. I first have to write the equation right? Is this the right equation? Cu(s)+HNO3(aq)-> Cu2+

abmon98 · Answer

For the reaction:HNO3(aq)+ Cu(s)->NO2(g)+ Cu(NO3)2(aq)+ H2O(l).  
The reducing half reaction is: H+(aq) + HNO3(aq) + e --> NO2(g) + H2O(l)
The oxidizing half reaction is: Cu(s) + 2NO3-(aq) -> Cu(NO3)2(aq) + 2e-
$$\times2 \to the reducing hald reaction$$
Cancel out the electrons so we end up with a reaction of 
Cu(s)+4HNO3(aq)-->2NO2(g)+2H2O(l)+Cu(NO3)(aq)