Question

Question #853bc

Answer 1

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: