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] ?

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

1. A

7.75 g

2. B

0.775 g

3. C

11g

4. D

10g

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### Solution:

Explanation :

$\begin{array}{l}2\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Ag}\text{\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{S}\text{\hspace{0.17em}\hspace{0.17em}}\to \text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{Ag}}_{2}\mathrm{S}\\ 2\text{\hspace{0.17em}\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}108\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{g}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{Ag}\text{\hspace{0.17em}}\mathrm{reacts}\text{\hspace{0.17em}}\mathrm{with}\text{\hspace{0.17em}}32\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{g}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{sulphur}\\ 10\text{\hspace{0.17em}}\mathrm{g}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{Ag}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{reacts}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{with}\text{\hspace{0.17em}}\frac{32}{216}\text{\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}10=\frac{320}{216}\text{\hspace{0.17em}}>\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\mathrm{g}\end{array}$

It means ‘S’ is limiting reagent

32 g of S reacts to form 216 + 32 = 248 g of ${\mathrm{Ag}}_{2}\mathrm{S}$

1 g of S reacts to form $\frac{248}{32}=7.75\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{g}$

Alternative method :

${n}_{eq}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}Ag\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{10}{108}=0.0925;$

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