Question

Question #fdf3f

Answer 1

Here's what I got.

Explanation:

The tricky part here is to make sure that you get the chemical formulas of the reactants and of the products right. Your two reactants are

• ammonium iodide
• lead(II) nitrate

Your two products are

• ammonium nitrate
• lead(II) iodide

You're dealing with four ionic compounds, which should tell you that this is a double replacement reaction. Therefore, you should look for one reactant to be insoluble and precipitate out of solution.

So, the chemical formulas for your compounds are

• `"NH"_4"I" ->` ammonium iodide
• `"Pb"("NO"_3)_2 ->` lead(II) nitrate
• `"NH"_4"NO"_3 ->` ammonium nitrate
• `"PbI"_2 ->` lead(II) iodide

Write the unbalanced chemical equation first

`"NH"_4"I"_text((aq]) + "Pb"("NO"_3)_text(2(aq]) -> "NH"_4"NO"_text(3(aq]) + "PbI"_text(2(s]) darr`

In this case, lead(II) iodide will be your insoluble solid that precipitates out of solution.

The quickest way to balance this chemical equation is to use the fact that three out of the four compounds are soluble in aqueous solution, which implies that they exist as ions.

Focus on the reactants. You can say that you have

`"NH"_4"I"_text((aq]) -> "NH"_text(4(aq])^(+) + "I"_text((aq])^(-)`

`"Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + color(red)(2)"NO"_text(3(aq])^(-)`

This means that the unbalanced chemical equation can be rewritten as

`"NH"_text(4(aq])^(+) + "I"_text((aq])^(-) + "Pb"_text((aq])^(2+) + color(red)(2)"NO"_text(3(aq])^(-) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-) +"PbI"_text(2(s]) darr`

The first thing to notice here is that you have `color(red)(2)` nitrate anions, `"NO"_3^(-)`, on the reactants side and only `1` on the products side.

To balance these anions out, multiply the compound that contains the nitrate anions on the products' side, i.e. ammonium nitrate, by `color(red)(2)`. This will give you

`"NH"_text(4(aq])^(+) + "I"_text((aq])^(-) + "Pb"_text((aq])^(2+) + color(red)(2)"NO"_text(3(aq])^(-) -> overbrace(color(red)(2)"NH"_text(4(aq])^(+) + color(red)(2)"NO"_text(3(aq])^(-))^(color(purple)("ammonium nitrate")) +` `"PbI"_text(2(s]) darr`

Now you have `color(red)(2)` ammonium ions, `"NH"_4^(+)`, on the products' side but only `1` on the reactants side. Moreover, you have `2` iodide anions on the products' side (stuck in the insoluble solid), but only `1` on the reactants' side.

This works out great because you need to multiply the compound that contains both these ions on the reactants' side, ammonium iodide, by `color(red)(2)`. You will thus have

`overbrace(color(red)(2)"NH"_text(4(aq])^(+) + color(red)(2)"I"_text((aq])^(-))^(color(blue)("ammonium iodide")) + "Pb"_text((aq])^(2+) + color(red)(2)"NO"_text(3(aq])^(-) -> overbrace(color(red)(2)"NH"_text(4(aq])^(+) + color(red)(2)"NO"_text(3(aq])^(-))^(color(purple)("ammonium nitrate")) +` `"PbI"_text(2(s]) darr`

Put the ions back together to get the overall balanced chemical equation

`color(green)(color(red)(2)"NH"_4"I"_text((aq]) + "Pb"("NO"_3)_text(2(aq]) -> color(red)(2)"NH"_4"NO"_text(3(aq]) + "PbI"_text(2(s]) darr`

The net ionic equation, which excludes spectator ions, will look like this

`color(red)(2)"I"_text((aq])^(-) + "Pb"_text((aq])^(2+) -> "PbI"_text(2(s]) darr`

Lead(II) iodide is a yellow insoluble solid that precipitates out of solution.