Lesson Plan
Advanced Chemistry
Grade 10 - STEM
I.
OBJECTIVES
At the end of this lesson, the students are expected to:
1. Define percent yield
2. Explain the importance of percent yield; and
3. Calculate percent yield
II.
SUBJECT MATTER
TOPIC: Percent Yield
REFERENCE:
https://www.khanacademy.org/science/ap-chemistrybeta/x2eef969c74e0d802:chemicalreactions/x2eef969c74e0d802:stoichiometry/a/limiting-reagents-andpercentyield#:~:text=Summary,much%20product%20can%20be%20formed.
https://www.thoughtco.com/definition-of-percent-yield-605899
III.
A.
B.
C.
D.
E.
PROCEDURES
Prayer
Greetings and Introduce as a Pre-service Teacher
Review
Recall the topic about limiting reactant.
Motivation
 You prepared cookie dough to make 5 dozen cookies. The phone
rings and you answer. While talking, a sheet of 12 cookies burn
and you have to throw them out. The rest of the cookies are okay.
 What is the percent yield of edible cookies?
F. Lesson proper
 Unlocking of terms
 Percent yield - the percent ratio of actual yield to the
theoretical yield

Teacher will introduce the formula of percent yield
G. Activity
What is the percent yield of CO when 30.0 g O2 are used? The actual
yield is 40.0 g CO.
2C(g) + O2(g)

2CO(g)
H. Analysis
When N2 and 5.00 g H2 are mixed, the reaction produces 16.0 g NH3.
What is the percent yield for the reaction?
N2(g) + 3H2(g) 
2NH3(g)
I. Abstraction
 Asked the students about the topic you discuss:
o How do you define percent yield?
o How important percent yield is?
o How will you solve the percent yields?
J. Application
 You burn 4.9 grams of magnesium in oxygen and find that actual
yield of magnesium oxide is 6.5 grams. What is the percent yield
of this reaction?
The equation for this reaction is:
2Mg + O2
IV.

2MgO
Evaluation

The student places an iron nail with a mass of 2. 32 g into a flask
of CuSO4. The nail reacts completely, leaving a quantity of copper
metal in the bottom of the flask. The student finds the mass of the
recovered copper to be 2.51 g. What is the expected yield? What
is the percent yield?
The equation for this reaction is:
Fe + CuSO4 
FeSO4 + Cu
V.
Assignment
2.50 g of hydrogen is reacted with 30.0 L of carbon monoxide at
STP.
CO(g) + 2 H2
CH3OH
2.50 g H2 produces 19.9 g CH3OH
30.0 L CO produces 42.9 g CH3OH
Prepared by:
Rhea T. Gigare - PST
Jessa Jhenne C. Javier - PST
Reviewed by:
Mrs. Jo Ann F. Cadayona
Cooperating Teacher