Balance the following redox reaction by inserting the appropriate coefficients. HNO3 + H2S --> NO + S + H2O

steps to balance redox reactions
(1) identify the oxidation number of every atom
(2) determine which atoms are oxidized and which are reduced
(3) write half reactions and include electrons
(4) balance electrons in half reactions
(5) combine balanced half reactions and cancel electrons
(6) combine and or add counter ions
(7) add any remaining species THEN balance them

*** 1 ***
H in H2S is +1
S in H2S is -2

H in HNO3 is +1
N in HNO3 is +5
O in HNO3 is -2

S in "S" is 0

N in NO = +2
O in NO = -2

*** 2 ***
N went from +5 to +2 and was REDUCED.. (reduction = reduction in charge)
S went from -2 to 0 and was OXIDIZED.. (oxidation is the opposite of reduction.. increase in charge

no atoms of H nor O changed charges

*** 3 ***
1 S(-2) ---> 1 S(0) + 2e's
1 N(+5) + 3 e's ---> 1 N(+2)

*** 4 ***
3x the first + 2x the second balances the e's

3 S(-2) ---> 3 S(0) + 6e's
2 N(+5) + 6 e's ---> 2 N(+2)

*** 5 ***
3 S(-2) + 2 N(+5) + 6 e's ---> 3 S(0) + 6e's + 2 N(+2)

3 S(-2) + 2 N(+5) ---> 3 S(0) + 2 N(+2)

*** 6 ***
3 H2S + 2 HNO3 ---> 3 S + 2 NO

*** 7 ***
notice you still have 3x2 + 2x1 = 8 H's on the left and NONE on the right?
notice you have 6 O's on the left and only 2 on the right?

what do you suppose 8H's and 6-2 = 4 O's make? that's right.. WATER so we need to add water on the right
3 H2S + 2 HNO3 ---> 3 S + 2 NO + __ H2O

and finally balancing the H's
3 H2S + 2 HNO3 ---> 3 S + 2 NO + 4 H2O