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

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.

i just learnt this recently but i need to look through my notes again