Question

In the lab you react 20g of potassium iodide with an excess of lead (II) nitrate to form 15g of lead (II) iodide precipitate. What is the percent yield of your experiment?

What is the balanced equation for this question?
What is the first ratio used to solve this question?
What is the second ratio used to solve this question?
What is the third ratio used to solve this question?
What is the percentage yield?

Answer 1

C

Step-by-step explanation:

Answer 2

The balanced equation for the reaction of potassium iodide with lead (II) nitrate to form lead (II) iodide is:
`$$2KI + Pb(NO_3)_2 \rightarrow 2KNO_3 + PbI_2$$`

To calculate the percent yield, we first need to calculate the theoretical yield. The molar mass of KI is approximately 166 g/mol and that of PbI2 is approximately 461 g/mol.

• The first ratio used is the stoichiometric ratio from the balanced equation, which is 2 moles of KI to 1 mole of PbI2.

• The second ratio is the conversion of grams of KI to moles using its molar mass. From 20 g of KI, we get approximately 0.12 moles of KI.

• The third ratio is the conversion of moles of PbI2 to grams using its molar mass. From 0.12 moles of PbI2, we get approximately 27.77 g of PbI2.

The percent yield is then calculated as the ratio of the actual yield (15 g) to the theoretical yield (27.77 g), multiplied by 100:

`$$\text{Percent Yield} = \left( \frac{15 \, \text{g}}{27.77 \, \text{g}} \right) * 100 = 54.02 \%$$`

So, the percent yield of the experiment is approximately 54.02%.