A) \[BaS{{O}_{4}}\]
B) \[SrS{{O}_{4}}\]
C) \[PbS{{O}_{4}}\]
D) \[CdS{{O}_{4}}\]
Correct Answer: C
Solution :
\[P{{b}^{+2}}\ +\ {{H}_{2}}S\ \xrightarrow[\text{acidic}]{}\ \underset{\text{(Black ppt}\text{.)}}{\mathop{PbS\downarrow \ }}\,+\ {{H}_{2}}\uparrow \] \[3PbS+8HN{{O}_{3}}\to 3Pb{{(N{{O}_{3}})}_{2}}+2NO+3S\]\[+4{{H}_{2}}O\] \[Pb{{(N{{O}_{3}})}_{2}}\ +\ {{H}_{2}}S{{O}_{4}}\ \to \underset{\text{(white ppt}\text{.)}}{\mathop{\ PbS{{O}_{4}}}}\,\downarrow \ +\ 2HN{{O}_{3}}\uparrow \]You need to login to perform this action.
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