Question

What is the mass of the precipitate formed when $50\text{mL}$ of $16.9\%$ solution of $AgN{O}_3$ is mixed with $50\text{mL}$ of $5.8\%$ $NaCl$ solution ? $(Ag=107.8,N=14,O=16,Na=23,Cl=35.5)$

• A. $7\text{g}$
• B. $14\text{g}$
• C. $28\text{g}$
• D. $3.5\text{g}$

Answer 1

The correct option is A $7\text{g}$

$16.9\text{g}AgN{O}_3$ is present in $100\text{mL}$ solution, $\therefore 8.45\text{g}AgN{O}_3$ is in $50\text{mL}$
$5.8\text{g}NaCl$ is present in $100\text{mL}$ solution, $\therefore 2.9\text{g}NaCl$ is present in $50\text{mL}$ solution
$\begin{array}{cccccccc}& AgN{O}_3& +& NaCl& \to & AgCl& +& NaN{O}_3\\ & \frac{8.45}{170}\text{mol}& & \frac{2.9}{58.5}& & & & \\ & =0.049\text{mol}& & =0.049\text{mol}& \to & 0& & 0\\ \text{after}& 0& & 0& \to & 0.049\text{mol}& & 0.049\text{mol}\\ \text{reaction}& & & & & & & \end{array}$
mass of $AgCl$ precipitated
$=0.049\times 143.5\text{g}$
$=7\text{g}AgCl$