Question

Question #e3eb3

Answer 1

Here's what I got.

Explanation:

The reaction template given to you is incomplete, so I assume that the problem wants you to figure out the two missing products.

In essence, you're dealing with the synthesis of potassium ferrioxalate trihydrate, `"K"_3"Fe"("C"_2"O"_4)_3 * 3"H"_2"O"`, by way of the reaction two hydrates

• `"FeCl"_3 * 6"H"_2"O" ->` iron(III) chloride hexahydrate
• `"K"_2"C"_2"O"_4 * "H"_2"O" ->` potassium oxalate monohydrate

In aqueous solution, you can write the two reactants as

`"FeCl"_3 * 6"H"_2"O"_text((aq]) -> "Fe"_text((aq])^(3+) + 3"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l])`

`"K"_2"C"_2"O"_4 * "H"_2"O"_text((aq]) -> 2"K"_text((aq])^(+) + "C"_2"O"_text(4(aq])^(2-) + "H"_2"O"_text((l])`

Without going into too much detail here, the oxalate anion, `"C"_2"O"_4^(2-)`, will act as a bidentate ligand and bind to the iron(III) cation, `"Fe"^(3+)`.

This will lead to the formation of the ferrioxalate complex ion, `["Fe"("C"_2"O"_4)_3]^(3-)`. The potassium cations will then act as a counterion and form the potassium ferrioxalate coordination complex, `"K"_3["Fe"("C"_2"O"_4)_3]`.

Notice that you need three potassium cations to balance the overall `3-` charge of the ferrioxalate complex anion.

This is an oversimplification of what happens because iron(II) chloride will actually hydrolyze in aqueous solution to form the hexaaquairon(III) complex cation, `["Fe"("H"_2"O")_6]^(3+)`. The oxalate anions will then replace water as the ligands to form the ferrioxalate complex anion.

So, rewrite the unbalanced chemical equation as

`"Fe"_text((aq])^(3+) + 3"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l]) + 2"K"_text((aq])^(+) + "C"_2"O"_text(4(aq])^(2-) + "H"_2"O"_text((l]) -> "K"_3["Fe"("C"_2"O"_4)_3] * 3"H"_2"O"_text((aq])`

Now, here is why I said that the template given to you is incomplete. This reaction will also produce aqueous potassium chloride, `"KCl"`, and additional water.

A complete picture of what goes on here would be

`"Fe"_text((aq])^(3+) + color(red)(3)"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l]) + 2"K"_text((aq])^(+) + "C"_2"O"_text(4(aq])^(2-) + "H"_2"O"_text((l]) -> "K"_3["Fe"("C"_2"O"_4)_3] * 3"H"_2"O"_text((aq]) + "K"_text((aq])^(+) + "Cl"_text((aq])^(-) + "H"_2"O"_text((l])`

Notice that you start with `2` potassium cations on the reactants' side, but you end up with `4` on the products' side.

Moreover, you start with `color(red)(3)` chloride anions on the reactants' side, and end up with `1` on the products' side. To balance the chloride anions, multiply the compound that contains them on the products' side, i.e. potassium chloride.

This will bring the total number of potassium cations present on the products' side to

`3 + color(red)(3) xx 1 = 6`

which means that you will have to multiply the potassium cations present on the reactants' side by `color(blue)(3)`.

But remember, you must multiply the compound that contains the potassium cations on the reactants' side. In this case, you must multiply the potassium oxalate monohydrate by `3`.

You will thus have

`"Fe"_text((aq])^(3+) + color(red)(3)"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l]) + color(blue)(3)overbrace((2"K"_text((aq])^(+) + "C"_2"O"_text(4(aq])^(2-) + "H"_2"O"_text((l])))^(color(blue)("potassium oxalate monohydrate")) -> "K"_3["Fe"("C"_2"O"_4)_3] * 3"H"_2"O"_text((aq]) + color(red)(3)overbrace(("K"_text((aq])^(+) + "Cl"_text((aq])^(-)))^(color(red)("potassium chloride")) + "H"_2"O"_text((l])`

This is equivalent to

`"Fe"_text((aq])^(3+) + color(red)(3)"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l]) + color(blue)(6)"K"_text((aq])^(+) + color(blue)(3)"C"_2"O"_text(4(aq])^(2-) + color(blue)(3)"H"_2"O"_text((l]) -> "K"_3["Fe"("C"_2"O"_4)_3] * 3"H"_2"O"_text((aq]) + color(red)(3)"K"_text((aq])^(+) + color(red)(3)"Cl"_text((aq])^(-) + "H"_2"O"_text((l])`

Now you're left with balancing the water molecules. You now have `9` water molecules on the reactants' side and `4` on the products' side. Multiply the additional water produced by the reaction by `6` to balance these out.

`"Fe"_text((aq])^(3+) + color(red)(3)"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l]) + color(blue)(6)"K"_text((aq])^(+) + color(blue)(3)"C"_2"O"_text(4(aq])^(2-) + color(blue)(3)"H"_2"O"_text((l]) -> "K"_3["Fe"("C"_2"O"_4)_3] * 3"H"_2"O"_text((aq]) + color(red)(3)"K"_text((aq])^(+) + color(red)(3)"Cl"_text((aq])^(-) + 6"H"_2"O"_text((l])`

Put all this back together to get the balanced chemical equation

`color(green)("FeCl"_color(red)(3) * 6"H"_2"O"_text((aq]) + color(blue)(3)"K"_2"C"_2"O"_4 * "H"_2"O"_text((aq]) -> "K"_3["Fe"("C"_2"O"_4)_color(blue)(3)] * 3"H"_2"O"_text((aq]) + color(red)(3)"KCl"_text((aq]) + 6"H"_2"O"_text((l]))`

Alternatively, you can write this as

`"FeCl"_(color(red)(3)text((aq])) + color(blue)(3)"K"_2"C"_2"O"_text(4(aq]) + 9"H"_2"O"_text((l]) -> "K"_3["Fe"("C"_2"O"_4)_color(blue)(3)] * 3"H"_2"O"_text((aq]) + color(red)(3)"KCl"_text((aq]) + 6"H"_2"O"_text((l])`

Crystals of potassium ferrioxalate trihydrate are green in color.