A) Bleaching powder
B) White vitriol
C) Mohr's salt
D) Microcosmic salt
Correct Answer: C
Solution :
\[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to \] \[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{H}_{2}}O+[O]\] \[\underset{\text{ }\!\![\!\!\text{ Mohr}-\text{salt }\!\!]\!\!\text{ }}{\mathop{2FeS{{O}_{4}}+}}\,{{H}_{2}}S{{O}_{4}}+[O]\to \] \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{H}_{2}}O]\times 5\] \[2KMn{{O}_{4}}+10FeS{{O}_{4}}+8{{H}_{2}}S{{O}_{4}}\to \] \[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+8{{H}_{2}}O\] Mohr-salt reducing agent \[KMn{{O}_{4}}/{{H}^{+}}\to \]oxidising agent
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