Question

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

• A. 10 g
• B. 0.77 g
• C. 11 g
• D. 7.75 g

Answer 1

The correct option is D 7.75 g

Identifying the limiting reagent:

Moles of $Ag=\frac{mass}{givenmass}=\frac{10}{108}=0.093mol$

Moles of $S=\frac{mass}{givenmass}=\frac{1}{32}=0.031mol$

Chemical reaction involved:

$2Ag+S\to A{g}_{2}S$
According to the reaction,
1 mole of S reacts with 2 mol of Ag.
Thus, 0.031 mol of S will react with 0.062 mol of Ag.

The given moles of Ag is 0.093 mol. This implies that Ag is present in excess, and S is the limiting reagent.
Thus, S will control the amount of $A{g}_{2}S$ formed.

Calculating the amount of $A{g}_{2}S$ formed:
According to the reaction,
1 mol of S produce 1 mol of $A{g}_{2}S$.
Thus, 0.031 mol of S will produce 0.031 mol of $A{g}_{2}S$.
Mass of $A{g}_{2}S$ formed = Number of moles × Molar mass
$=0.031mol\times 248gmo{l}^{-1}$

$=7.69g$
Hence, the amount of $A{g}_{2}S$ formed is 7.69 g.

So, option (a) is the correct answer.