Question

Someone pleas help I have a test tomorrow so please answer this as soon as you see this. I have 30.0 grams of potassium Iodide with 24.0 grams of lead nitrate. The products of the reaction are lead iodide and potassium nitrate. What is the limiting reactant? What is the mass of the excess reactant left over?

If you dry the lead iodide and find it's mass to be 32.5 g what is the percent yield?

Again please help I have tried everything.

Answer 1

Let's walk through this together, and pay close attention. The answers are bolded and underlined but I insist you to read through the whole thing and understand what's being applied here.

Write down the balanced equation first:

`Pb(NO_(3))_(2) \ + \ 2KI \ ----- > \ PbI_2 \ + \ 2KNO_(3)`

24g 30g ???

To find the limiting reactant, convert both of the masses to moles and the lowest one is the limiting reagent.

Molar mass of Lead (II) Nitrate: 331.2g/mol

Molar mass of Potassium Iodide: 166g/mol

Remember:
`moles \ = \ (mass)/(molar \ mass)`

`(24)/(331.2) = 0.0724 \ mol\ of\ Lead \ (II) \ Nitrate\\\\(30)/(166) = 0.180 \ mol \ of \ Potassium \ Iodide`

As you can see here, the compound with the lower mol value is Lead (II) Nitrate, meaning Lead (II) Nitrate is our limiting reactant.

To find the excess reactant which is the other reactant present, Potassium Iodide, you must do the mol to mol ratio of Lead (II) Nitrate to Potassium Iodide (Basically the numbers in front/coefficients of the compound in the balanced equation)

The ratio is 1 : 2, meaning you have to multiply the amount of mol of Lead (II) Nitrate by 2 to get the mol of Potassium Iodide.

`0.0724 \ * \ 2 = \ 0.1448 \ mol \ of \ Potassium \ iodide`

Now we change the mol into mass by multiplying it by the molar mass of potassium iodide

`0.1448 \ * \ 166 = \ 24.04g \ of \ Potassium \ iodide`

To get the amount of excess reactant, you have to subtract the original mass amount you had which was 30g by the mass you just got which is 24.04g

30 - 24.04 = 5.96g of Potassium Iodide is in excess left over

Our final question is to calculate the percentage yield of lead iodide, which we can do so by dividing the actual yield produced by the theoretical yield, and multiplying the answer of that by a 100 to get the percent.

We have our experimental yield which is 32.5g but we don't have our theoretical. We have to calculate it using our limiting reactant and mol ratios. The ratio of Lead (II) Nitrate to Lead (II) Iodide is 1 : 1 so all we have to is to change 0.0724 mol into the mass value by multiplying the mol with the molar mass of Lead (II) Iodide.

Molar mass of Lead (II) Iodide: 461.01g/mol

`0.0724 \ * \ 461.01 = \ 33.4g`

So the theoretical yield is 33.4g and the actual yield was 32.5g

`(32.5)/(33.4) * 100 = 97.3 \ percent`

The percentage yield of Lead (II) Iodide is 97.3%