8 g of sulphur is burnt to form $C l_{2}$ which is oxidized by $C l_{2}$ water. The solution is treated with $B a C l_{2}$ solution. Number of moles of $B a S O_{4}$ precipitated is

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No. of moles of BaSO4 precipitates when reagent [Cu(NH3)4]SO4 treated with excess of BaCl2 :-

The correct option is C 1[Cu(NH3)4]SO4+BaCl2⟶[Cu(NH3)4]Cl2+BaSO41 mole The number of a mole of BaSO4 precipitates With reagent BaCl2 (excess). Hence 1 mole BaS04.Correct Answer - C––

No. of moles of $B a S O_{4}$ precipitates when reagent $[C u (N H_{3})_{4}] S O_{4}$ treated with excess of $B a C l_{2}$ :-

The correct option is C $1$
$[C u {({N H}_{3})}_{4}] {S O}_{4} + {B a C l}_{2} ⟶ [C u {({N H}_{3})}_{4}] {C l}_{2} + {B a S O}_{4}      1 mole$
The number of a mole of BaSO4 precipitates With reagent BaCl2 (excess).
Hence 1 mole BaS04.
Correct Answer - $C - -$