Question

What weight of AgCl will be precipitated when a solutuion containing 4.77g of NaCl is added to solution of 5.77g ${\mathrm{AgNO}}_3$?

Answer 1

$\mathrm{NaCl}+{\mathrm{AgNO}}_3\to \mathrm{AgCl}+\mathrm{Na}\left({\mathrm{NO}}_3\right)$
MW 58.5g 170g 143.5g
Mole 1 1 1
No. of moles of NaCl = $\frac{4.77}{58.5}=0.08\mathrm{mol}$
No. of moles of ${\mathrm{AgNO}}_3$ = $\frac{5.77}{170}=0.033\mathrm{mol}$
Here ${\mathrm{AgNO}}_3$is limiting reagent.
$\because$ 1 mol of ${\mathrm{AgNO}}_3$ gives = 1 mol of AgCl
$\therefore$ 0.033 mol of ${\mathrm{AgNO}}_3$ gives = 0.033 mol of AgCl
So, weight of AgCl formed = 0.033 $\times$ 143.5 = 4.735g