Question

$50.0mL$ of $0.10M$ sodium chloride solution is mixed with $25.0mL$ of $0.25M$ silver nitrate solution. What mass of silver chloride should form from this reaction?

Answer 1

Given, $M_{NaCl}=0.1M,V_{NaCl}=50mL$

$M_{AgN{O}_3}=0.25M,V_{AgN{O}_3}=25mL$

$\Rightarrow$ No. of millimoles of $NaCl=\text{Molarity}\times \text{Volume of}NaCl=0.1\times 50=5mMoles$

$\Rightarrow$ No. of millimoles of $AgN{O}_3=\text{Molarity}\times \text{Volume of}AgN{O}_3=0.25\times 25=6.25mMoles$

So, the limiting reagent is $NaCl.$

So, it produces $5millimoles$ of $AgCl.$

$\Rightarrow$ Mass of $AgCl$ produced $=(5\times {10}^{-3})\left(143.5\right)=0.7175grams$