One method proposed for removing $S O_{2}$ from the flue gases of power plants involves the reaction of $S O_{2}$ with aqueous $H_{2} S .$ How many moles of sulphur are formed by passing $S O_{2}$ obtained by burning 1.0 ton (one ton) of coal containing 3.5% sulphur into aqueous solution of $H_{2} S$

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answer is 410.1.

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Detailed Solution

$\begin{array}{l} Mass of sulphur in 1ton= \frac{3.5 \times 1000 \times 1000}{100} = 35000 g m s \\ S + O_{2} \to S O_{2} \\ S O_{2} + H_{2} S \to 2 H_{2} O + 3 S \\ F o r e a c h 32 g m o f s u l p h u r i n c o a l 96 g o f s u l p h u r i s p r e c i p i t a t e d . \\ T h e n 35, 000 g m o f s u l p h u r i s 35000 \times 3 = 105000 g m s u l p h u r e x i s t s S_{8} \\ a s m o l e c u l e s . S o t h e n u m b e r o f m o l e o f s u l p h u r f o r m e d i s \frac{105000}{256} = 410 m o l e s \end{array}$