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] ?

• A. 7.75 g
• B. 0.775 g
• C. 11 g
• D. 10 g

Answer 1

Answer: (A)

7.75 g

Answer 2

(1)

(1) Explanation : 2 Ag + S $\rightarrow$ Ag2 S

2 × 108 g of Ag reacts with 32 g of sulphur

10 g of Ag reacts with $\frac{32}{216}$ × 10 = $\frac{320}{216}$> 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 $= \frac{248}{32}$ = 7.75 g

Alternately

neq of Ag=$\frac{10}{108}$= 0.0925 neq of S =$\frac{1}{16}$ = 0.0625 (neq

= number of equivalents)

Since neq of S is less than neq of Ag

$\Rightarrow$ 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