What is the mass of the precipitate formed when 50 mL of 16.9 %  solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution? (Ag=107.8, N=14,O=16,Na=23,Cl=35.5)

16.9% solution of AgNO3 means 16.9 g of AgNO3 in 100 mL of solution.16.9 g of AgNO3 in 100 mL solution = 8.45 g of AgNO3 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 :                      AgNO3+NaCl⟶AgCl+NaNO3Initially       8.45/170        2.9/58.5                =  0.049 mol    = 0.049 molFinally            0                        0           0.049 mol    0.049 mol∴  Mass of AgCl precipitated =0.049×143.3                                                                         =7.02≈7 g