Question

The addition of $0.15\text{moles}$ of $NaCl$ to an aqueous solution containing $0.05\text{moles}$ of $AgN{O}_3$ results in the formation of an insoluble white precipitate of $AgCl$. Find the limiting reagent and the amount of $AgCl$ that gets formed.
$(Ag=108\text{u},Cl=35.5\text{u})$

• A. $NaCl$, $7.175\text{g}$
• B. $AgN{O}_3$, $7.175\text{g}$
• C. $NaCl$, $14.350\text{g}$
• D. $AgN{O}_3$, $14.350\text{g}$

Answer 1

The correct option is B $AgN{O}_3$, $7.175\text{g}$

The corresponding reaction is:
$NaCl\left(aq\right)+AgN{O}_3\left(aq\right)\to NaN{O}_3\left(aq\right)+AgCl\left(s\right)$
Moles of $NaCl=0.15\text{mol}$
Moles of $AgN{O}_3=0.05\text{mol}$
From the balanced stoichiometric equation,
$1$ mole of $NaCl$ reacts with $1$ mole of $AgN{O}_3$
$0.15$ moles of $NaCl$ will react with $0.15$ 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 $AgCl$
$0.05$ moles of $AgN{O}_3$ will produce $0.05$ moles of $AgCl$
Amount of precipitate formed
$=\text{moles of}AgCl\text{formed}\times \text{Molar mass of}AgCl$
$=(0.05\times 143.5)\text{g}$
$=7.175\text{g}$