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?

Options

A. 28g
B. 3.5g
C. 7g
D. 14g

Answer

C

Explanation

16.9g `AgNO_(3)` is present in 100 mL solution.
`therefore 8.45 g AgNO_(3)` is present in 50 mL solution.
5.8g NaCl is present in 100 mL solution
5.8 g NaCl is present in 100 mL solution
`therefore` 2.9 g NaCl is present in 50 mL solution
`AgNO_(3) + NaCl to AgCl + NaNO_(3)`
`(8.45)/(170) ` mol `(2.9)/(58.5)`
`= 0.049` mol = 0.049 mol `to 0 " " 0`
after `" "0 " "0 " " to ` 0.049 mol 0.049 mol
reaction
mass of AgCl precipitated
= 0.049 `xx` 143.5g = 7g AgCl.