What is the mass of the precipitate formed when 50 mL of 16.9% solution of `AgNO_(3)` is mixed with 50 mL of 5.8% NaCl solution?

Explanation

100 mL of `AgNO_(3)` solution have mass = 16.9 g
50 mL of `AgNO_(3)` solution have mass = 8.45 g
Number of moles of `AgNO_(3)` present `= ("Mass of" AgNO_(3))/("Molar mass")=((8.45g))/((170"g mol"^(-1)))=0.048` mol
100 mL of NaCl solution have mass = 5.8 g
50 mL of NaCl solution have mass = 2.9 g
Number of moles of NaCl present `= ("Mass of NaCl")/("Molar mass")=((2.9g))/((58.5"g mol"^(-1)))=0.049` mol
The chemical equation for the reaction may be written as :
`{:(,AgNO_(3)+,NaClrarr,AgCl("ppt")+,NaNO_(3)),("Initial moles","0.049 mol","0.049 mol",0,0),("Final moles",0,0,"0.049 mol",0.049 mol"):}`
Number of moles of AgCl formed `= 0.049` mol
Mass of AgCl formed `= (0.049 mol xx 143.5 g mol^(-1))`
`= 7.03 ~~ 7.0 g`.