LIMITING REACTANTS AND THE AMOUNT OF PRODUCTS FORMED
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Physical Science
OBJECTIVES
After going through this lesson, you are expected to:
1. Determine the limiting reactant/reagent in a chemical reaction,
2. Understand the excess reactant/regent in a chemical reaction, and
3. Calculate the amount of product/s formed in a chemical reaction.;
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Review
____ 1. A catalyst is a substance that can be added to a reaction to increase the reaction rate without getting consumed in the process.
Directions: clap once if the statement is true. Clap twice If the statement is false.
____ 2. Heat is not a catalyst since it can’t be taken out of the reaction in the same amount it was inputted into the reaction.
Directions: clap once if the statement is true. Clap twice If the statement is false.
___ 3. Catalysts increase the rates of chemical reactions by raising the activation energy of the reactions.
Directions: clap once if the statement is true. Clap twice If the statement is false.
____ 4. Collisions only result in a reaction if the particles collide with a certain maximum energy called the activation energy of the reaction.
Directions: clap once if the statement is true. Clap twice If the statement is false.
____5. Decreasing the concentration of reactants increases the collision frequency between reacting particles.
Directions: clap once if the statement is true. Clap twice If the statement is false.
Let me know which
of the following is limiting item :
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Example of Limiting reactant:
How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter?
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With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.
Four table settings can be made.
What is the limiting item?
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The limiting item is the spoon.
| Initially | Use | Left over |
plates | 5 | 4 | 1 |
forks | 6 | 4 | 2 |
spoons | 4 | 4 | 0 |
knives | 7 | 4 | 3 |
In a table setting, there is 1�plate, 1 fork, 1 knife, and�1 spoon. �
If a carpenter had two tabletops and seven table legs, how many table/ tables he could build?
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He could only build one four-legged table. The number of table legs is the limiting factor.
Pre- test
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Identify the key term referred in each statement. Write your answer on a separate sheet of paper. Choose your answer from the words above
______1. Substances originally present in a chemical reaction that are used to produce products after the reaction.
______2. Substance produced in the chemical reaction.
_______3. A written chemical symbol to represent the process of chemical reaction.
_______4. In a chemical reaction, it limits the amount of substance that may be produced.
________5. A reactant which is not used up after the chemical reaction is done.
Product Limiting reactant Reactant Chemical equation
Excess reactant
Pre- test Part II
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6. It is the reactant that produces a lesser amount of product.
c. Limiting Reactant d. Co-Factor
7. What is the calculation of the relationship of reactant and product in chemical equation.
a. Molality b. Stoichiometry
c. Balancing Equation d. Percent Yield
8. It is the maximum amount of product you would expect from a reaction based on the amount of limiting reagent
c. Excess Reactant d. Product Difference
Pre- test
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9. . They are the reactants that are not used up when the reaction is finished.
c. Limiting Reagents d. Solution
10. It deals with the process that involves rearrangement of the molecular or ionic structure of a substance to form a new substance or product.
a. Chemical Equilibrium b. Chemical Reaction
c. Chemical Symbols d. Stoichiometry
KEY TERMS:
Stoichiometry
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is a branch of chemistry that establishes mole and mass relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.
LIMITING REACTANT
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A limiting reactant in a chemical reaction is the substance that
-Is used up first.
-Stops the reaction.
-Limits the amount of product that can form.
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A substance's molar mass is calculated by multiplying its relative atomic mass by the molar mass constant (1g/mol).
The molar mass constant can be used to convert mass to moles.
By multiplying a given mass by the molar mass, the amount of moles of the substance can be calculated.
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CHEMICAL REACTION
CHEMICAL BALANCING
CHEMICAL BALANCING
CHEMICAL BALANCING
CHEMICAL BALANCING
QUIZ TIME
COPY AND ANSWER
SHOW YOUR SOLUTION ON A
1 WHOLE SHEET OF PAPER
Balancing Chemical Equations
(1) __ Na₃PO₄ + __ HCl → __ NaCl + __ H₃PO₄
(2) __ TiCl₄ + __ H₂O → __ TiO₂ + __ HCl
(3) __ C₂H₆O + __ O₂ → __ CO₂ + __ H₂O
(4) __ FeS + __ O₂ → __ Fe₂O₃ + __ SO₂
(5) __ CaCl₂ + __ Na₃PO₄ → __ Ca₃(PO₄)₂ + __ NaCl
ARE YOU’VE DONE!!!
EXAMPLE PROBLEM #1:
STEP 1: WRITE THE BALANCED CHEMICAL EQUATION
FOR THE REACTION DESCRIBE IN THE PROBLEM.
STEP 2: CALCULATE THE MOLAR MASS OF THE REACTANTS.
STEP 3: Divide the calculated molar mass of the given element or compound from the given mass(given in problem).
STEP 4: Then, the number of moles are divided by the coefficient indicated in the balanced chemical equation.
For Al: 0.167 mol of Al X 1 mol of AlCl3
1 mol of Al
= 0.167 mol of AlCl3
For NaCl: 0.039 mol of NaCl X 1 mol of AlCl3
3 mol of NaCl
= 0.013 mol of AlCl3
STEP 5: Find the grams of AlCl3 formed.
We can use the moles of limiting reagent plus the stoichiometric ratios from our balanced reaction to calculate the theoretical yield.
Theoretical Yield = 0.013 mol of AlCl3 x 133.33g of AlCl3
1 mol of AlCl3
= 1.73 g of AlCl3
The theoretical yield is usually expected to have units of mass, so we can convert moles of AlCl3 to grams using the molecular weight:
Atomic mass Al - 26.98
Atomic mass Cl - 3x 35.45 = 106.35
106.35 + 26.98 = 133.33g
EXAMPLE PROBLEM #2:
If a reaction vessel contains 0.15 mol of KO2 and 0.10 mol of H2O, what is the limiting reactant? How many moles of oxygen can be produced?
KO2 + H20 KOH + O2
STEP 1: WRITE THE BALANCED CHEMICAL EQUATION FOR THE REACTION DESCRIBE IN THE PROBLEM.
4KO2 + 2H20 4KOH + 3O2
STEP 2: The number of moles indicated in the given problem times the number of mol of the element or compound being asked in the problem divided by the coefficient indicated in the balanced chemical equation.
For KO2: 0.15 mol of KO2 x 3 mol of O2
4 mol of KO2
= 0.113 mol of O2
For H20: 0.10 mol of H2O x 3 mol of O2
2 mol of H2O
= 0.15 mol of O2
0.15 mol of KO2 and 0.10 mol of H2O
4KO2 + 2H20 4KOH + 3O2
The Limiting Reactant – KO2
Amount of O2 formed – 0.113 mol
Note: The amount of product that can be formed based on the limiting reactant is called the THEORETICAL YIELD.
LET'S EXERCISE
What is the limiting reactant if we start with 2.80g of Al and 4.25 g of Cl?
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2.80 g
26.98 g/mol
=
0. 1037 mol
Cl =
4.25 g
35. 45g/mol
= 0.1199 mol
R1
R2
Al =
LET'S EXERCISE
What is the limiting reactant if we start with 2.80g of Al and 4.25 g of Cl?
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0. 1037 mol of Al X
Cl =
2 mol AlCl3
3 mol Cl2
0.1199 mol Cl X
R1
R2
Al =
2 mol Al
2 mol AlCl3
=
0. 1037 mol of AlCl3
= 0. 0799 mol of AlCl3
The limiting reactant is Al.
Cl is the excess reactant
(0.1199 mol)
LET'S EXERCISE
What is the limiting reactant if we start with 3.20g of C and 1.25 g of O2?
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2C + O2 (g) 2 CO2 (s)
3.20 g
12. 01g/mol
=
0. 266 mol
1.25 g
32.00 g/mol
O2=
= 0.039 mol
R1
R2
C=
LET'S EXERCISE
What is the limiting reactant if we start with 3.20g of C and 1.25 g of O2?
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2C + O2 (g) 2 CO2 (s)
2 mol CO2
2 mol C
=
0. 266 mol C
1 mol O2
O2=
0.039 mol O2
R1
R2
C=
X
0.266 mol of CO2
X
2 mol CO2
=0.078 mol of CO2
The limiting reactant is Oxygen gas (O2). (0.039 mol)
Carbon (C) is the excess reactant - (0.266 mol)
LET'S EXERCISE
A 2.00 g sample of ammonia is mixed with 4. 00 g of oxygen . Which is the limiting reactant?
1. Write a balanced equation.
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4NH3 (g) + 5 O2 4NO(g) + 6H2O(g)
2.00g NH3 X
1 mol NH3
17.0 g NH3
X
4 mol NO
4 mol NH3
X
30.0 g NO
1 mol NO
=
3.53 g NO
R1
R2
LET'S EXERCISE
A 2.00 g sample of ammonia is mixed with 4. 00 g of oxygen . Which is the limiting reactant?
1. Write a balanced equation.
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4NH3 (g) + 5 O2 4NO(g) + 6H2O(g)
4.00g O2 X
1 mol O2
32.0 g O2
X
4 mol NO
5 mol O2
X
30.0 g NO
1 mol NO
=
3.00 g NO
The limiting reactant is O2 (oxygen gas)
Post- test
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Limiting Reactant And The Amount Of Product Form
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BF3 (g) + H2 (g) B + HF
Limiting Reactant And The Amount Of Product Form
2. How many moles of NaCl are produced from the reaction between 20.0g CuCl2 and 15.0 g Na NO3, as outlined in the reaction below?
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CuCl2 + NaNO3 (NO3)2+ NaCl
3. The reaction for the combustion of Ethane is shown by the balanced reaction below.
C2H6 + O2 H2O + CO2
1.45g C2H6 was burned in the presence of 4. 50g O2. Determine the limiting reactant.
THANK YOU!
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