CuS + NO₃¯ ---> NO + Cu²⁺ + HSO₄¯

In this reaction, CuS is a solid substance with a nitric acid solution being poured on it. Chemically, both the Cu and the S must be accounted for, but ONLY at the end, in the final answer.

In this problem, we can eliminate the Cu from both sides of the equation and make things a bit simpler. The reason we can do this has to do with what happens to the Cu during the reaction. Note that it is a +2 ion in solution as a product. However, what is its charge in the CuS? The answer is +2, so the Cu was neither reduced nor oxidized in the reaction. That means we can eliminate it during the balancing and act like only S²¯ is in solution. However, we must add the Cu back in at the end.

So, let's the drop the Cu to get:

S²¯ + NO₃¯ ---> NO + HSO₄¯

the two half-reactions are as follows:

balancing them gives:

make the number of electrons equal:

the not so final answer:

put the Cu²⁺ back in.