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 of `N_(2)` and 32 g of `O_(2)`
D. 3.5 g

Answer

A

Explanation

16.9% `AgNO_(3)` solution means that 16.9 g of `AgNO_(3)` is present in 100 mL of solution.
50 mL of solution contains `AgNO_(3) = 16.9/100 xx 50`
= 8.45 g
`=8.45/170 = 0.049` mol
`AgNO_(3) + NaCl to AgCl + NaNO_(3)`
1 mol of `AgNO_(3)` combines with 1 mol of NaCl to give 1 mol of AgCl.
0.049 mol of `AgNO_(3)` combines with 0.049 mol of NaCl to give AgCl
0.049 mol
`=0.049 xx 143.5= 7g`