Someone tell me how to balance the equation by hit and trial method...

1. 3H2S + 2HNO3 > 2NO + 4H2O + 3S

Explanation:

H2S + HNO3 >  NO + H2O + S

There are 3 O's on left and  only 2 on right so  put 2 before NO:

H2S + HNO3 > 2NO + H2O + S

Now we have 2 N's on left so put 2 before HNO3:

H2S + 2HNO3 > 2NO + H2O + S

4 H's on the left so

H2S + 2HNO3 > 2NO + 2H2O + S

That makes 6 O's on left so  we write

H2S + 2HNO3 > 2NO + 4H2O + S

Now there 8 H's on the right but only 4 on the right so we adjust H2S and S:

3H2S + 2HNO3 > 2NO + 4H2O + 3S

Now it balances!

Basically you 'play around' with the  quantities of each element in this way.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons