A) nitrate
B) nitrite
C) bromide
D) iron
Correct Answer: A
Solution :
At the junction of salt formation and \[\text{FeS}{{\text{O}}_{\text{4}}}\]solution with cone. \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]a brown ring is obtained \[\text{(FeS}{{\text{O}}_{4}}\text{.NO)}\text{.}\] \[\underset{\begin{smallmatrix} \text{nitrate}\,\text{salt} \\ \text{soluble}\,\text{in}\,\text{water} \end{smallmatrix}}{\mathop{NO_{3}^{-}+}}\,6FeS{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\] \[+2NO+3{{H}_{2}}O\] \[FeS{{O}_{4}}+NO\to \underset{brown\,ring}{\mathop{FeS{{O}_{4}}.NO}}\,\]
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