How to balance the reaction equation between copper (I) sulfide and nitric acid?

Question

What is the sum of stoichiometric coefficients of redox reaction: $\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$?

This exercise is from a multiple choice exam. I have balanced the equation: $$\ce{1 Cu2S + 4 HNO3 -> 1 S + 2 NO2 + 1 Cu(NO3)2 + 2 H2O.}$$

Therefore I have calculated the sum of coefficients as $\sum a = 11$. This has not been offered as an answer and the correct answer is 20.

Answer 1

$$\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$$

Let's balance this reaction using the half-reaction method.

The oxidation half reaction (O.H.R) would be:

$$\ce{\overset{+1}{Cu}_2\overset{-2}{S} -> 2Cu^2+ + S^0 + 4e-}\tag{O.H.R}\label{ohr}$$

The reduction half reaction (R.H.R) here is:

$$\ce{H\overset{+5}{N}O3 +e- -> \overset{+4}{N}O2} \tag{R.H.R}\label{rhr}$$

The final reaction would be $4\times\eqref{rhr} + 1\times\eqref{ohr}$.

Therefore the reaction would be partially balanced as:

$$\ce{Cu2S + 4HNO3 -> 2 Cu^2+ + S + 4NO2}$$

Now, we have $\ce{Cu(NO3)2}$ and not $\ce{Cu^2+}$, so we need to add the 4 more nitrate ions which implies 4 more molecules of $\ce{HNO3}$

So now, we have the equation as:

$$\ce{Cu2S + 8HNO3 -> 2 Cu(NO3)2 + S + 4NO2}$$

Now, we balance the hydrogens and oxygens, the left has 8 hydrogens and 24 oxygens, whereas the right has no hydrogens and 20 oxygens. Therefore, we need to add 4 water molecules to the right. Therefore, the final reaction becomes:

$$\ce{Cu2S + 8 HNO3 -> S + 4 NO2 + 2Cu(NO3)2 + 4 H2O}$$

Now, $\sum a = 1 + 8 + 1 + 4 + 2 + 4 = 20$ and not 23.