Question

A chemistry student is performing a lab exercise where they are reacting aqueous Lead (II) Nitrate with aqueous Potassium Iodide. The student begins the reaction with 10.0 g of Lead (II) Nitrate and 12.0 g of Potassium Iodide. At the end of the experiment the student has collected 11.4 grams of the bright yellow solid precipitate Lead (II) Iodide.
Write the full balanced equation. Include the appropriate states of matter.
How many atoms of Lead are found in the product collected by the student?
Based upon the information above, what is the percent yield of this student’s lab?

Answer 1

Answer:  Pb(NO3)2(aq) + 2KI(aq)  —> PbI2(s) + 2KNO3(aq)

1.57x10^22 atoms of Pb

87% yield

Explanation:

10.0 g Pb(II) nitrate = 10/333 moles = 0.03 moles

12.0 g KI = 12/166 moles = 0.072 moles (0.06 moles required)

11.4 g PbI2 = 11.4/421 moles = 0.026 moles

expected yield 0.03 moles, yield = 87%

atoms in 11.4 g = 0.026 moles = 0.026*6.02214076*10^23 atoms = 1.57*10^22