Question #853bc

1 Answer
Dec 11, 2014

The answer is #0.12g#.

Essentially, what you are dealing with is a double replacement precipitation reaction in which two soluble ionic compounds (two soluble salts) react to form an insoluble precipitate.

Starting from the balanced chemical equation, we get

#NaCl_((aq)) + AgNO_(3(aq)) -> NaNO_(3(aq)) + AgCl_((s))#

The complete ionic equation, which better describes the reaction since ionic compounds dissociate into ions when dissolved in water, is

#Ag_((aq))^(+) + NO_(3(aq))^(-) + Na_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s)) + Na_((aq))^(+) + NO_(3(aq))^(-)#

Notice that #Na_((aq))^(+)# and #NO_(3(aq))^(-)# did not participate in the reaction since they can be found on both sides of the equation -> such ions are called spectator ions.

This gives us the net ionic equation, which shows us what ions participate in the reaction that forms the precipitate

#Ag_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s))#

Now, we know that we have a #1:1# mole ratio for #NaCl# and #AgNO_3#; the number of moles of #AgNO_3# can be determined from molarity, #C = n_(solute)/(V_(solution)#

#n_(solute) = C * V_(solution) = (0.100 mol es)/(L) * (20 * 10^(-3)) L# = 0.0020 moles

This means that the number of moles of #NaCl# must be 0.0020 as well. Knowing #NaCl#'s molar mass - #58.5 g/(mol)# - we get

#m_(NaCl) = 0.0020 mol es * 58.5 g/(mol) = 0.12g#

This is true because the number of moles of #NaCl# dissociate into moles of #Na^+# and moles of #Cl^(-)#, the same being true for #AgNO_3#; therefore, the number of #NaCl# moles must be at least equat to the number of #AgNO_3# moles.

A more in depth analysis could be done using the concentrations of the compounds, the reaction's coefficient (#Q_(sp)#), and #K_(sp)#.

Here's a video of the reaction: