Question

Consider this reaction: H2S + HNO3 S + NO + H2O Reduction half-reaction: N+5 + 3e- N+2 Oxidation half-reaction: S-2 S + 2e- After balancing the chemical equation, what are the coefficients for these compounds?

Answer 1

Answer: 1. 3

2. 2

Step-by-step explanation:

Answer 2

Coefficient of
`H_2S` = 3

Coefficient of
`HNO_3` = 2

Coefficient of
`NO` = 2

Coefficient of
`H_2O` = 4

Coefficient of
`S` = 3

Step-by-step explanation:

Overall reaction

`H_2S + HNO_3\rightarrow S + NO + H_2O`

Reduction half-reaction:

`N^(+5) + 3e^-\rightarrow N^(+2)`...(1)

Oxidation half-reaction:

`S^[-2}\rightarrow S + 2e^-`...(2)

2×(1) + 3×(2)

`2N^(+5) + 6e^-+3S^[-2}\rightarrow 2N^(+2)+3S + 6e^-`

`2N^(+5) +3S^[-2}\rightarrow 2N^(+2)+3S`

`3H_2S + 2HNO_3\rightarrow 3S + 2NO + 4H_2O`

Coefficient of
`H_2S` = 3

Coefficient of
`HNO_3` = 2

Coefficient of
`NO` = 2

Coefficient of
`H_2O` = 4

Coefficient of
`S` = 3