Mole ConceptHard
Question
If 10 g of Ag reacts with 1 g of sulphur, the amount of Ag2S formed will be [Atomic weight of Ag = 108, S = 32] ?
Options
A.7.75 g
B.0.775 g
C.11 g
D.10 g
Solution
(A) Explanation : 2 Ag + S → Ag2 S
2 × 108 g of Ag reacts with 32 g of sulphur
10 g of Ag reacts with
× 10 = 
> 1 g
It means ′S′ is limiting reagent
32 g of S reacts to form 216 + 32 = 248 g of Ag2S
1 g of S reacts to form =
= 7.75 g
Alternately
neq of Ag =
= 0.0925 neq of S = 
= 0.0625 (neq = number of equivalents)
Since neq of S is less than neq of Ag
⇒ 0.0625 eq of Ag will react with 0.0625 eq of S to form 0.0625 eq of Ag2S
Hence, amount of Ag2S = neq × Eq. wt. of Ag2S
= 0.0626 × 124 = 7.75 g
2 × 108 g of Ag reacts with 32 g of sulphur
10 g of Ag reacts with
× 10 = > 1 gIt means ′S′ is limiting reagent
32 g of S reacts to form 216 + 32 = 248 g of Ag2S
1 g of S reacts to form =
= 7.75 gAlternately
neq of Ag =
= 0.0925 neq of S = = 0.0625 (neq = number of equivalents)Since neq of S is less than neq of Ag
⇒ 0.0625 eq of Ag will react with 0.0625 eq of S to form 0.0625 eq of Ag2S
Hence, amount of Ag2S = neq × Eq. wt. of Ag2S
= 0.0626 × 124 = 7.75 g
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