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?

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Akinny Akinny · Answer 1 · 2019-11-08T11:58:54+01:00

Answer:

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