Question

In the lab you produce and collect 15.46 g of KNO3. If you carried out the reaction with 43.65 g of Pb(NO3)2 and 32.92 g of Kl what is your percent yield?Round to two decimal placesPb(NO3)2 + 2KI - Pbl2 + 2KNO3

Answer 1

Answer:

The percent yield is 77.18%.

Explanation:

1st) According to the balanced reaction, 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 2 moles of KNO3 and 1 mole of PbI2.

We can convert the moles to grams using the molar mass of Pb(NO3)2 (331.2g/mol), KI (166g/mol) and KNO3 (101g/mol):

- Pb(NO3)2 conversion: 331.2g

- KI conversion: 332g

- KNO3 conversion: 202g

2nd) Now, with the stoichiometry of the reaction in grams we can calculate the grams of KNO3 that must be produced (this is the Theoretical yield):

$\begin{gathered} 332gKI-202gKNO_3 \\ 32.92gKI-x=\frac{32.92gKI*202gKNO_3}{332gKI} \\ x=20.03gKNO_3 \end{gathered}$

So, this is the theoretical amount of KNO3.

3rd) With the theoretical yield (20.03g) and the Actual yield (15.46g) we can calculate the percent tield:

$\begin{gathered} \text{ Percent yield=}\frac{\text{ Actual yield}}{\text{ Theoretical yield}}*100 \\ \text{ Percent yield=}\frac{15.46g}{20.03g}*100 \\ \text{ Percent yield=77.18\%} \end{gathered}$

So, the percent yield is 77.18%.