Question

What weight of $AgCl$ (in g) will be precipitated when a solution containing 1.99 g of $NaCl$ is added to a solution containing 5.78 g of $AgN{O}_3$?
(Molar masses of $Ag,Cl$ and $Na$ are $108,35.5$ and 23 g/mol respectively)

Answer 1

$\begin{array}{cccccccc}& AgN{O}_3& +& NaCl& \to & AgCl& +& NaN{O}_3\\ \text{mol mixed:}& \frac{5.78}{170}& & \frac{1.99}{58.5}& & & & \\ & 0.034& & 0.034& & 0& & 0\\ \text{mol left:}& 0& & 0& & 0.034& & 0.034\end{array}$

moles of $AgCl$ formed = 0.034 mol
amount of $AgCl$ formed = $0.034mol\times 143.5g/mol$
$W_{AgCl}$ = 4.88 g