Question

The addition of $0.1\text{mole}$ of $KBr$ to an aqueous solution containing $0.05\text{moles}$ of $AgN{O}_3$ results in the formation of an insoluble white precipitate of $AgBr$. Find the amount of $AgBr$ that gets formed.
$(Ag=108\text{u},Br=80\text{u})$

• A. $9.4\text{g}$
• B. $4.7\text{g}$
• C. $7.8\text{g}$
• D. $8.2\text{g}$

Answer 1

The correct option is A $9.4\text{g}$

The corresponding reaction is:
$KBr\left(aq\right)+AgN{O}_3\left(aq\right)\to KN{O}_3\left(aq\right)+AgBr\left(s\right)$
Moles of $KBr=0.1\text{mol}$
Moles of $AgN{O}_3=0.05\text{mol}$
From the balanced stoichiometric equation,
$1$ mole of $KBr$ reacts with $1$ mole of $AgN{O}_3$
$0.1$ moles of $KBr$ will react with $0.1$ moles of $AgN{O}_3$
So, $AgN{O}_3$ is the limiting reagent.
As per the reaction, $1$ mole of $AgN{O}_3$ produces $1$ mole of $AgBr$.
$0.05$ moles of $AgN{O}_3$ will produce $0.05$ moles of $AgBr$.
Amount of precipitate formed
$=\text{moles of}AgBr\text{formed}\times \text{Molar mass of}AgBr$
$=(0.05\times 188)\text{g}$
$=9.4\text{g}$