A) \[PbS{{O}_{4}}\]
B) \[PbSi{{O}_{3}}\]
C) Both [a] and [b]
D) \[CaSi{{O}_{3}}\]
Correct Answer: C
Solution :
By intermittently filled lime the blown air in the sinterer gets diffused and \[PbS\] particles come under limited supply of air therefore formation of \[PbS{{O}_{4}}\] is prevented. \[PbS+\frac{3}{2}{{O}_{2}}\to PbO+S{{O}_{2}}\] \[PbO+\underset{gangue}{\mathop{Si{{O}_{2}}}}\,\to PbSi{{O}_{3}}\] \[PbSi{{O}_{3}}+\underset{more\,basic}{\mathop{CaO}}\,\to CaSi{{O}_{3}}+\underset{less\,basic}{\mathop{PbO}}\,\]
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