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

Answer

1

Explanation

`16.9%` solution of `AgNO_(3)` implies that every 100 mL of solution contain `16.9 gAgNO_(3)`. Thus, `50 mL` of solution will contain `8.45 g AgNO_(3)`.
Similarly, `5.8% NaCl` solution implies that every `100 mL` of solution contains `5.8 g NaCl`. Thus, `50 mL` of solution will contain `2.9 g NaCl`.
`n_(AgNO_(3))=(mass_(AgNO_(3)))/("molar mass"_(AgNO_(3)))`
`=(8.45 g)/(170 g mol^(-1))=0.049 mol`
`n_(NaCl)=(mass_(NaCL))/("molar mass"_(NaCl))`
`=(2.9 g)/(5.85" g "mol^(-1))=0.049 mol`
According to the balanced chemical equation
`AgNO_(3)(aq)+NaCl(aq) rarr AgCl(s)+NaNO_(3)(aq)`

`Mass_(AgCl)" precipitated"=(n//AgCl)("molar mass"_(AgCl))`
`=(0.049 mol)(143.5 g mol^(-1))`
`=7.0 g`