Question

$50.0mL$ of $0.10M$ sodium chloride solution is mixed with $25.0mL$ of $0.25M$ silver nitrate solution. After filtering and drying the product, $0.20g$ of silver chloride was produced.
What is the percent yield from this reaction?

Answer 1

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

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

$AgN{O}_3+NaCl\to AgCl\downarrow +NaN{O}_3$

No. of millimoles of $NaCl=M_{NaCl}\times V_{NaCl}=5mMoles$

No. of millimoles of $AgN{O}_3=M_{AgN{O}_3}\times V_{AgN{O}_3}=6.25mMoles$

So, $AgCl$ formed is $5mMoles$

Mass of $AgCl$ formed $(5\times {10}^{-3})\left(143\right)=0.715g$

$\%$ Yield $=\frac{0.2}{0.715}\times 100=27.97\%$