When h2s(g) is bubbled into aqueous nitric acid solutions, s(s), no2(g), no(g), n2(g), and nh4+ (aq) are all formed.
(a) (8 pts) write the four balanced chemical reactions (using only electrically neutral species) that describe this
process.
(b) (4 pts) for each reaction in part (a), evaluate the number of moles of electrons transferred per mole of
h2s(g).

Explanation:

a. The reactions produced when H₂S is bubbled into nitric acid solutions are:

3H₂S + 2HNO₃ → 2NO(g) + 3S(s) + 4H₂O

H₂S + HNO₃ → NO₂(g) + S(s) + H₂O

4H₂S + HNO₃ → NH₄OH + 4S(s) + 2H₂O (NH₄⁺ in aqueous solution produce NH₄OH)

5H₂S + 2HNO₃ → N₂(g) + 5S(s) + 6H₂O

In all reactions the nitrogen is oxidized and sulfur is reduced. The problem says the sulfur is reduced in just S(s) and the four ways in which nitrogen is oxidized.

b. - 3H₂S + 2HNO₃ → 2NO(g) + 3S(s) + 4H₂O

2HNO₃ → 2NO The number of electrons transferred are:

2×N:+5 → 2×N:+2; +10 → +4 = 6 electrons, as there are 3 moles of H₂S the moles of electrons transferred per mole of H₂S are 6/3 = 2 moles e⁻

- H₂S + HNO₃ → NO₂(g) + S(s) + H₂O

HNO₃ → NO₂ The number of electrons transferred are:

N:+5 → N:+4; +5 → +4 = 1 electron, as there is 1 mole of H₂S the moles of electrons transferred per mole of H₂S are 1/1 = 1 mole e⁻

- 4H₂S + HNO₃ → NH₄OH + 4S(s) + 2H₂O

HNO₃ → NH₄OH The number of electrons transferred are:

N:+5 → N:-3; +5 → -3 = 8 electrons, as there are 4 moles of H₂S the moles of electrons transferred per mole of H₂S are 8/4 = 2 moles e⁻

- 5H₂S + 2HNO₃ → N₂(g) + 5S(s) + 6H₂O

2HNO₃ → 2N₂ The number of electrons transferred are:

2×N:+5 → 2×N:0; +10 → 0 = 10 electrons, as there are 5 moles of H₂S the moles of electrons transferred per mole of H₂S are 10/5 = 2 moles e⁻

I hope it helps!