Let a,b,c,d,e,f be the number of molecules of each reactant. Th echecmical eqaution becomes
aCuS+bHNO_3->cCu(NO_3)_2+dS+eH_2O+fNO
Obsrvation reveals that nmber of molecules of H NO_3 must be an even number to accommodate H_2 on the right of the equation.
Let's start balancing for the radical NO_3.
On the right hand side of the equation the radical is either part of
atom Cu(NO_3)_2 or can be obtained from combination of H_2O and NO. We may choose lowest integer 1 for either and double the number of atoms of the other. However, choosing lowest integer 1 for NO will rquire chosing 2 atoms of H_2O. This will increase the number of atoms of H required for balancing. Therefore, choosing 2 atoms of NO or f=2, the equation becomes
aCuS+bHNO_3->cCu(NO_3)_2+dS+eH_2O+2NO
With this we need 3 more atoms of O for the second atom of N. Implies e=4. The equation becomes
aCuS+bHNO_3->cCu(NO_3)_2+dS+4H_2O+2NO
Now after having fixed number of atoms of H on the right side, it follows that b must=8. To give us
aCuS+8HNO_3->cCu(NO_3)_2+dS+4H_2O+2NO
Now balacaing 24 number of O atoms of left side with the right side atoms we obtain c=3. This fixes a=3=d. We obtain the balanced equation as
3CuS+8HNO_3->3Cu(NO_3)_2+3S+4H_2O+2NO
