Question

For the reaction Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃, how many moles of lead iodide are produced from 189.9 g of potassium iodide?

a. 0.572
b. 4.38e4
c. 1.24
d. 2.48

Answer 1

To find the moles of lead iodide produced from 189.9 g of potassium iodide (KI), calculate the moles of KI using its molar mass, use the mole ratio from the balanced equation to find the moles of PbI2 produced, and convert the moles of PbI2 to grams using the molar mass of PbI2.

Step-by-step explanation:

Given the balanced equation:

Pb(NO3)2 + 2KI → PbI2 + 2KNO3

To find the moles of lead iodide produced from 189.9 g of potassium iodide (KI), we need to use the molar mass of KI:

1. 
   
3. Calculate the moles of KI using its molar mass (39.10 g/mol).
4. 
   
6. Use the mole ratio from the balanced equation to find the moles of PbI2 produced.
7. 
   
9. Convert the moles of PbI2 to grams using the molar mass of PbI2 (461.01 g/mol).

By following these steps, we find that the moles of lead iodide produced from 189.9 g of potassium iodide is approximately 0.572 moles.