Answer:
3 Cu + 8 HNO3 --> 3 Cu(NO3)2 + 2 NO + 4 H2O
Explanation:
Make sure both sides are equal
3 Cu + 8 HNO3 --> 3 Cu(NO3)2 + 2 NO + 4 H2O
// start by those elements that change their oxidation degree
Cu and N
// also you can write reduction-oxidation reactions
[tex]Cu^{0}[/tex] + 2 [tex]e^{-}[/tex] --> [tex]Cu^{-2}[/tex] | 2
[tex]N^{+5}[/tex] - 3 [tex]e^{-}[/tex] --> [tex]N^{+2}[/tex] | 3
// write the numbers of electrons that are lost/gained as the coefficients of the opposite elements
// then check if H and O are the same on both sides
// adjust if they aren't.
