Question

In the lab you react 23 g of potassium iodide with an excess of lead (II) nitrate to form 18 g of lead (II) iodide precipitate. What is the percent yield of your experiment?

A) 28
B) 56
C) 84
D) 98

Answer 1

B) Percent yield = 56%

Step-by-step explanation:

Given data:

Mass of potassium iodide = 23 g

Mass of lead iodide formed = 18 g

Percent yield = ?

Solution:

Chemical equation:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

Number of moles of potassium iodide:

Number of moles = mass / molar mass

Number of moles = 23 g/ 166 g/mol

Number of moles = 0.14 mol

Now we will compare the moles of PbI₂ and KI:

KI : PbI₂

2 : 1

0.14 : 1/2×0.14 = 0.07

Theoretical yield of PbI₂:

Mass = number of moles × molar mass

Mass = 0.07 × 461 g/mol

Mass = 32.27 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 18 g/ 32.27 g × 100

Percent yield = 56%