Question

A 1.0 L sample of an aqueous solution contains 0.10 mol of NaCl and 0.10 mol of CaCl2. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the Cl- as AgCl(s) ? (Assume that AgCl is insoluble.)

Answer 1

Answer:

The answer to your question is: 0.3 moles of AgNO₃

Explanation:

1.0 L sample

0.1 mol of NaCl

0.1 mol of CaCl₂

AgNO₃ = ? moles

Reactions

               NaCl + AgNO₃ ⇒ AgCl + NaNO₃

Then                1 NaCl mol ---------------  1 AgNO₃

                     0.1 mol           --------------    x

             x = 0.1 moles of AgNO₃ needed

             

              CaCl₂ + 2 AgNO₃ ⇒ 2 AgCl + Ca(NO₃)₂

Then                 1 mol of CaCl₂ ------------- 2 moles of AgNO₃

                       0.1 mol              -------------      x

            x = 0.2 moles of AgNO₃

Total moles of AgNO₃ = 0.1 + 0.2 = 0.3

Answer 2

0.30 mol $AgNO_3$ to precipitate all of the $Cl^{-}$ ions.

The first step would be to write out the equation as follows:-  

$NaCl(aq) + CaCl_2(aq) + AgNO_3(aq) -> AgCl(s) + NaNO_3(aq) + Ca(NO_3)_2(aq)$

Now we need to balance the reaction:

$NaCl(aq) + CaCl_2(aq) + 3AgNO_3(aq) -> 3AgCl(s) + NaNO_3(aq) + Ca(NO_3)_2(aq)$

So from the above balanced reaction, we can see that for every mole of NaCl and every mole of $CaCl_2$, we need 3 moles of $AgNO_3$  to precipitate all of the $Cl^{-}$ ions.

So,  0.10 mol of KCl and 0.10mol of $CaCl_2$ , we need a total of 0.30mol  to precipitate all of the $Cl^{-}$ ions.

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