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Q.
If $10 \,g$ of $Ag$ reacts with $1 \, g$ of sulphur, the amount of $Ag_2S$ formed will be [Atomic weight of $Ag = 108, S =32$]?
Some Basic Concepts of Chemistry
Solution:
Explanation$: 2 Ag + S \to Ag_2 S$
$2 \times 108 g$ of $Ag$ reacts with $32 g$ of sulphur
$10 g$ of $Ag$ reacts with$\frac{32}{216}\times10=\frac{320}{216}>1 g$
It means $'S'$ is limiting reagent
$32 g$ of $S$ reacts to form $216 + 32 = 248 g$ of $Ag_2S$
$1 g$ of $S$ reacts to form $\frac{248}{32}=7.75 g$
Alternately $n_{eq}$ of $Ag=\frac{10}{108}=0.0925;$
$n_{eq}$ of $S=\frac{1}{16}=0.0625$ ($n_{eq}$ = number of equivalents)
Since $n_{eq}$ of $S$ is less than $n_{eq}$ of $Ag$
$\Rightarrow 0.0625 eq$ of $Ag$ will react with $0.0625 eq$ of $S$ to form $0.0625 eq$ of $Ag_2S$
Hence, amount of $Ag_2S = n_{eq}\times Eq. wt.$ of $Ag_2S$
$= 0.0626 \times 124 = 7.75 g$