Question

Write the balanced chemical reacton between 20.0mL of .10M lead(II) nitrate and 30.0mL of .20M potassium iodide?

a) Which reactant is the limiting reactant?
b) What was the mass of the precipitate?
c) What are the concentrations of the ions still floating around when the reaction is complete?
(All potassium salts are soluble)

Thank you so much for helping!

Answer 1

Here's what I got.

Explanation:

!! LONG ANSWER !!

Aqueous lead(II) nitrate, `"Pb"("NO"_3)_2`, will react with aqueous potassium iodide, `"KI"`, to form lead(II) iodide, `"PbI"_2`, an insoluble ionic compound that precipitates out of solution, and aqueous potassium nitrate, `"KNO"_3`.

The balanced chemical equation that describes this reaction looks like this

`"Pb"("NO"_ 3)_ (2(aq)) + color(red)(2)"KI"_ ((aq)) -> "PbI"_ (2(s)) darr + 2"KNO"_ (3(aq))`

The two reactants are consumed in a `1:color(red)(2)` mole ratio, so look to use this mole ratio to find the limiting reagent.

Use the molarities and volumes given to you to calculate how many moles of each reactant are delivered to the reaction

`20.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.10 moles Pb"("NO"_3)_2)/(1color(red)(cancel(color(black)("L"))))`

`= 2.0 * 10^(-3)"moles Pb"("NO"_3)_2`

`30.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.20 moles KI")/(1color(red)(cancel(color(black)("L"))))`

`= 6.0 * 10^(-3)"moles KI"`

So, do you have enough moles of lead(II) nitrate to make sure that all the moles of potassium iodide take part in the reaction?

`2.0 * 10^(-3) color(red)(cancel(color(black)("moles Pb"("NO"_3)_2))) * (color(red)(2)color(white)(a)"moles KI")/(1color(red)(cancel(color(black)("mole Pb"("NO"_3)_2))))`

`= 4.0 * 10^(-3)"moles KI"`

You have `6.0 * 10^(-3)` moles of potassium iodide, but you only need `4.0 * 10^(-3)` moles to make sure that all the moles of lead(II) nitrate react `->` lead(II) nitrate will be the limiting reagent, i.e. it will be completely consumed before all the moles of potassium iodide can get the chance react.

`color(green)(|bar(ul(color(white)(a/a)color(black)("Pb"("NO"_3)_2 -> "limiting reagent")color(white)(a/a)|)))`

In other words, potassium iodide is in excess. After the reaction is complete, you will be left with

`6.0 * 10^(-3)"moles KI" - 4.0 * 10^(-3)"moles KI"`

`= 2.0 * 10^(-3)"moles KI" " " " "color(orange)("( * )")`

Keep this in mind.

Now, lead(II) nitrate is produced in a `1:1` mole ratio by lead(II) nitrate. You know that all the moles of lead(II) nitrate are consumed by the reaction, which means that the reaction produces

`2.0 * 10^(-3) color(red)(cancel(color(black)("moles Pb"("NO"_3)_2))) * "1 mole PbI"_2/(1color(red)(cancel(color(black)("mole Pb"("NO"_3)_2))))`

`= 2.0 * 10^(-3)"moles PbI"_2`

To convert moles of lead(II) iodide to grams, use the compound's molar mass

`2.0 * 10^(-3) color(red)(cancel(color(black)("moles PbI"_2))) * "461.01 g"/(1color(red)(cancel(color(black)("mole PbI"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.92 g")color(white)(a/a)|)))`

The answer is rounded to two sig figs.

In order to find the concentrations of the ions that are still floating around in solution, i.e. of the spectator ions, write the complete ionic equation

`"Pb"_ ((aq))^(2+) + color(blue)(ul(color(black)(2"NO"_ (3(aq))^(-)))) + color(blue)(ul(color(black)(color(red)(2)"K"_ ((aq))^(+)))) + color(red)(2)"I"_ ((aq))^(-) -> "PbI"_ (2(s)) darr + color(blue)(ul(color(black)(2"K"_ ((aq))^(+)))) + color(blue)(ul(color(black)(2"NO" _(3(aq))^(-))))`

As you can see, the nitrate anions and the potassium cations act as spectator ions.

For lead(II) nitrate, you know that every mole that dissolves in aqueous solution produces `2` moles of lead(II) nitrate. This means that you have

`2.0 * 10^(-3)color(red)(cancel(color(black)("moles NO"_3^(-)))) xx "2 moles NO"_3^(-)/(1color(red)(cancel(color(black)("mole NO"_3^(-)))))`

`= 4.0 * 10^(-3)"moles NO"_3^(-)`

For potassium iodide, you know that every mole that dissolves in aqueous solution produces `1` mole of potassium cations. This means that you have

`6.0 * 10^(-3)color(red)(cancel(color(black)("moles KI")))* "1 mole K"^(+)/(1color(red)(cancel(color(black)("mole KI"))))`

`= 6.0 * 10^(-3)"moles K"^(+)`

Now backtrack to the fact that potassium iodide is in excess. According to `color(orange)("( * )")`, you have an excess of `2.0 * 10^(-3)` moles of potassium iodide.

These moles will produce

`2.0 * 10^(-3) color(red)(cancel(color(black)("moles KI"))) * "1 mole I"^(-)/(1color(red)(cancel(color(black)("mole KI")))`

`= 2.0 * 10^(-3)"moles I"^(-)`

Therefore, you can say that after the reaction is complete, the solution will contain

• `2.0 * 10^(-3)"moles I"^(-)`
• `4.0 * 10^(-3)"moles NO"_3^(-)`
• `6.0 * 10^(-3)"moles K"^(+)`

The total volume of the solution will be

`V_"total" = "20.0 mL" + "30.0 mL" = "50.0 mL"`

The concentration of the ions will thus be

`["I"^(-)] = (2.0 * color(red)(cancel(color(black)(10^(-3))))"moles")/(50.0 * color(red)(cancel(color(black)(10^(-3))))"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.040 M")color(white)(a/a)|)))`

`["NO"_3^(-)] = (4.0 * color(red)(cancel(color(black)(10^(-3))))"moles")/(50.0 * color(red)(cancel(color(black)(10^(-3))))"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.080 M")color(white)(a/a)|)))`

`["K"^(+)] = (6.0 * color(red)(cancel(color(black)(10^(-3))))"moles")/(50.0 * color(red)(cancel(color(black)(10^(-3))))"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.12 M")color(white)(a/a)|)))`

The answers are rounded to two sig figs.