Question

If $10g$ of $Ag$ reacts with $1g$ of sulphur, the amount of $A{g}_{2}S$ formed will be [Atomic weight of $Ag=108,S=32$]?

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

Answer 1

Answer: (A)

Explanation$:2Ag+S\to A{g}_{2}S$
$2\times 108g$ of $Ag$ reacts with $32g$ of sulphur
$10g$ of $Ag$ reacts with$\frac{32}{216}\times 10=\frac{320}{216}>1g$
It means ${}^{\prime }{S}^{\prime }$ is limiting reagent
$32g$ of $S$ reacts to form $216+32=248g$ of $A{g}_{2}S$
$1g$ of $S$ reacts to form $\frac{248}{32}=7.75g$
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 $Ag$
$\Rightarrow 0.0625eq$ of $Ag$ will react with $0.0625eq$ of $S$ to form $0.0625eq$ of $A{g}_{2}S$
Hence, amount of $A{g}_{2}S=n_{eq}\times Eq.wt.$ of $A{g}_{2}S$
$=0.0626\times 124=7.75g$