MCQ
When ${H_2}S$ is passed through acidified $KMn{O_4},$ we get
- A${K_2}S{O_3}$
- B$Mn{O_2}$
- C$KHS{O_3}$
- ✓$Sulphur$
✓
Answer
Correct option: D.
$Sulphur$
d
(d) $2KMn{O_4} + 3{H_2}S{O_4} + 5{H_2}S\, \to \,$
(d) $2KMn{O_4} + 3{H_2}S{O_4} + 5{H_2}S\, \to \,$
${K_2}S{O_4} + 2MnS{O_4} + 8{H_2}O + 5S$
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