Silver ions can be precipitated from aqueous solutions by the addition of aqueous chloride: Ag + (aq) + Cl - (aq) → AgCl (s) Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.366 M AgNO3 solution to completely precipitate the silver?

0.5352 grams ≈ 0.54 grams

Explanation:

The stoichiometry of the equation is given below:

AgNO3 (aq) + NaCl (aq) ⇒ AgCl (s) + NaNO3 (aq)

1 mole 1 mole 1 mole 1 mole

The mole ratio between the reactants AgNO3 and NaCl is 1 : 1

From the question, the number of moles of AgNO3 used can be calculated using the formula below:

n = CV/1000

where n = number of moles

C = concentration in mol/dm3

V = Volume in mL

n (AgNO3) = 25 x 0.366/1000

= 0.00915‬ moles

From the equation of reaction, I mole of AgNO3 reacts with 1 mole of NaCl; so 0.00915 moles of AgNO3 will react with exactly 0.00915 moles of NaCl.

To covert moles of NaCl to grams, we use the formula below:

Number of moles (n) = Mass in grams (m) / Molar mass (M)

n = m/M

n = 0.00915 moles

m = ?

M = Molar mass of NaCl [23 + 35.5 = 58.5 grams]

m = n x M

= 0.00915 x 58.5

= 0.5353 grams

≈ 0.54 grams