How many grams of solid NaCl must be added to 30.0 mL of 0.255 M AgNO3 solution to completely precipitate the silver?

Question

How many grams of solid NaCl must be added to 30.0 mL of 0.255 M AgNO₃ solution to completely precipitate the silver?

Answer 1

Write the balanced equation:

NaCl(aq) + AgNO3(aq) ==> NaNO3(aq) + AgCl(s)

Find moles AgNO3 present: 0.0300 L x 0.255 mol/L = 0.00765 moles AgNO3

From the balanced equation 1 mol AgNO3 react with 1 mole NaCl, so you need 0.00765 moles NaCl

To find grams of NaCl, simply convert moles NaCl to grams NaCl:

0.00765 moles NaCl x 58.5 g NaCl/mole = 0.4475 g = 0.448 g NaCl (to 3 significant figures)