Question

Consider the reaction.

2Pb(s)+O2(g)⟶2PbO(s)

An excess of oxygen reacts with 451.4 g of lead, forming 318.8 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

Percent Yield = 65.57%

Step-by-step explanation:

To find the percent yield, you need to
(1) convert grams Pb to moles (using the atomic mass of Pb), then
(2) convert moles Pb to moles PbO (using the mole-to-mole ratio from the balanced equation), then
(3) convert moles PbO to grams (using the molar mass of PbO), and then (4) calculate the percent yield.
It is important to arrange the conversions in a way that allows for the cancellation of units.

Steps 1 - 3

Atomic Mass (Pb): 207.20 g/mol

Molar Mass (PbO): 207.20 g/mol + 15.999 g/mol = 223.199 g/mol

2 Pb(s) + O₂(g) -----> 2 PbO(s)
^ ^

451.4 g Pb 1 mole Pb 2 moles PbO 223.199 g
----------------- x -------------------- x ----------------------- x ---------------------- =
207.20 g 2 moles Pb 1 mole PbO

= 486.2 g PbO

Step 4

The actual yield is determined by running an experiment, whereas the theoretical yield is determined using stoichiometry and a balanced equation.

Actual Yield = 318.8 g PbO

Theoretical Yield = 486.2 g PbO

actual yield
Percent Yield = ------------------------------ x 100%
theoretical yield

318.8 g PbO
Percent Yield = ------------------------ x 100%
486.2 g PbO

Percent Yield = 65.57%