How to balance the reaction equation between copper (I) sulfide and nitric acid?
$$\ce{Cu2S + HNO3 -&gt; S + NO2 + Cu(NO3)2 + H2O}$$
Let's balance this reaction using the half-reaction method.
The oxidation half reaction (O.H.R) would be:
$$\ce{\overset{+1}{Cu}_2\overset{-2}{S} -&gt; 2Cu^2+ + S^0 + 4e-}\tag{O.H.R}\label{ohr}$$
The reduction half reaction (R.H.R) here is:
$$\ce{H\overset{+5}{N}O3 +e- -&gt; \overset{+4}{N}O2} \tag{R.H.R}\label{rhr}$$
The final reaction would be $4\times\eqref{rhr} + 1\times\eqref{ohr}$.
Therefore the reaction would be partially balanced as:
$$\ce{Cu2S + 4HNO3 -&gt; 2 Cu^2+ + S + 4NO2}$$
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:
$$\ce{Cu2S + 8HNO3 -&gt; 2 Cu(NO3)2 + S + 4NO2}$$
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:
$$\ce{Cu2S + 8 HNO3 -&gt; S + 4 NO2 + 2Cu(NO3)2 + 4 H2O}$$
Now, $\sum a = 1 + 8 + 1 + 4 + 2 + 4 = 20$ and not 23.