What is the mass of the precipitate formed when 50 mL of 16.9 %  solution of ${\mathrm{AgNO}}_3$ is mixed with 50 mL of 5.8% NaCl solution? 

$(\mathrm{Ag}=107.8, \mathrm{N}=14,\mathrm{O}=16,\mathrm{Na}=23,\mathrm{Cl}=35.5)$

16.9% solution of ${\mathrm{AgNO}}_3$ means 16.9 g of $AgN{O}_3$ in 100 mL of solution.

16.9 g of $AgN{O}_3$ in 100 mL solution = 8.45 g of ${\mathrm{AgNO}}_3$ in 50 mL solution. 

Similarly, 5.8% of NaCl in 100 mL solution = 2.9 g of NaCl in 50 mL solution.

The reaction can be represented as :

                      ${\mathrm{AgNO}}_3+\mathrm{NaCl}⟶\mathrm{AgCl}+{\mathrm{NaNO}}_3$

Initially       8.45/170        2.9/58.5

                =  0.049 mol    = 0.049 mol

Finally            0                        0           0.049 mol    0.049 mol

$\therefore \text{ Mass of }\mathrm{AgCl}\text{ precipitated }=0.049\times 143.3$

                                                                         $=7.02\approx 7 \mathrm{g}$