Copy of Theoretical Yield, Limiting Reagent Percent Yield Practice 1-2 (5/28/2026)

Last updated about 2 hours ago

5 questions

Note from the author:

Instructions

Untitled Section

Practice Questions

Copy of Theoretical Yield, Limiting Reagent Percent Yield Practice 1-2 (5/28/2026)

Last updated about 2 hours ago

5 questions

Note from the author:

C.4.5 Use a balanced chemical equation to calculate the quantities of reactants needed and products made in a chemical reaction that goes to completion.

Learning Goals:
I can use a balanced chemical equation to calculate the quantities of reactants needed in a completed chemical reaction.
I can use a balanced chemical equation to calculate the products made in a completed chemical reaction.

Instructions

C.4.5 Use a balanced chemical equation to calculate the quantities of reactants needed and products made in a chemical reaction that goes to completion.

Learning Goals:
I can use a balanced chemical equation to calculate the quantities of reactants needed in a completed chemical reaction.
I can use a balanced chemical equation to calculate the products made in a completed chemical reaction.

Untitled Section

Limiting Reagent (Reactant)
Reactant 1 + Reactant 2 → Product(s)
Usually, one of the reactants is fully consumed in a reaction before the other.
Limiting Reagent - the reactant that is fully consumed, first.
The reaction will stop when the limiting reagent is consumed.

Limiting Reagent (cont - 2)
Example - Grilled Cheese Sandwiches
2 slices of bread + 1 slice of cheese → 1 grilled cheese
Cheese is consumed, first.
Cheese is the limiting reagent.

Limiting Reagent (cont - 3)
Identifying the Limiting Reagent:
Sometimes we will find this via stoichiometric calculations.
In many of your first questions, it will tell you one of the reactions is in excess and another has a specific quantity; usually the reactant in excess is NOT the limiting reagent while the reactant with a provided quantity is the limiting reagent.

Yield Calculations - three aspects
Theoretical Yield - calculated
Actual Yield - measured in a lab
Percent Yield - calculated

Theoretical Yield 
Amount of product that may be produced.
Use stoichiometric calculations of a balanced equation.
Expressed in mass (grams) or moles.

Actual Yield
Amount of product that is produced.
Measured amount in a lab.
Commonly less than the theoretical yield.
Expressed in mass (grams) or moles.

Percent Yield
The degree that the theoretical yield is achieved.
Calculated using a ratio of the actual yield compared to the theoretical yield.
Expressed as a percent.

Example 1
CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.
 

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.

Set-up the equation (3 parts)
Mass of reactant to moles of reactant (need the molar mass of reactant).

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.

Set-up the equation (3 parts)
Mass of reactant to moles of reactant.
Moles of reactant to moles of product (1 mol of CuSO4 → 1 mol Cu).

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.

Set-up the equation (3 parts)
Mass of reactant to moles of reactant.
Moles of reactant to moles of product.
Moles of product to mass of product (need the molar mass of product).

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.

Set-up the equation (3 parts)
Mass of reactant to moles of reactant.
Moles of reactant to moles of product.
Moles of product to mass of product.
 

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.
Theoretical Yield = 0.507 g Cu

Step 3 - Percent Yield
Actual Yield = 0.392 g Cu

CuSO4 + Zn → Cu + ZnSO4
1.274 g of CuSO4 was reacted with an excess of Zn to produce 0.392 g Cu.
Step 1 – Balance the Equation

Step 2 - Theoretical Yield
Need molar masses of CuSO4 and Cu
Use periodic table.
Theoretical Yield = 0.507 g Cu

Step 3 - Percent Yield
Actual Yield = 0.392 g Cu

Example 2
CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

Step 2 - Theoretical Yield
Need molar mass of CCl4 and CF2Cl2
Use periodic table.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

Step 2 - Theoretical Yield
Need molar mass of CCl4 and CF2Cl2
Use periodic table.

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

Step 2 - Theoretical Yield
Need molar mass of CCl4 and CF2Cl2
Use periodic table.


Set-up the equation (3 parts)
Mass of reactant to moles of reactant (need the molar mass of reactant.)

CCl4 + HF → CF2Cl2 + HCl
32.9 g CCl4 is reacted with an excess of HF to produce 12.5 g CF2Cl2
Step 1 - Balance the Equation.

Step 2 - Theoretical Yield
Need molar mass of CCl4 and CF2Cl2
Use periodic table.


Set-up the equation (3 parts)
Mass of reactant to moles of reactant (need the molar mass of reactant.)

Example 3 - Limiting Reagent
Si + N2 → Si3N2
2.00 g Si was reacted with 1.50 g N2 to form 1.52 g Si3N2.
In this example, we will be using both reactants and the product to determine the limiting reagent.

Si + N2 → Si3N2
2.00 g Si was reacted with 1.50 g N2 to form 1.52 g Si3N2.
Step 1 - Balance the Equation.

Si + N2 → Si3N2
2.00 g Si was reacted with 1.50 g N2 to form 1.52 g Si3N2.
Step 1 - Balance the Equation.

Si + N2 → Si3N2
2.00 g Si was reacted with 1.50 g N2 to form 1.52 g Si3N2.
Step 1 - Balance the Equation.

Si + N2 → Si3N2
2.00 g Si was reacted with 1.50 g N2 to form 1.52 g Si3N2.
Step 1 - Balance the Equation.

Step 2 - Theoretical Yield.
Need molar mass of both reactants and the product (necessary to deter. limiting reagent).
Use periodic table

Si + N2 → Si3N2
2.00 g Si was reacted with 1.50 g N2 to form 1.52 g Si3N2.
Step 1 - Balance the Equation.

Step 2 - Theoretical Yield.
Molar masses

Practice Questions

Practice Questions
You Need:
Scientific Calculator (recommend - Natural Scientific Calculator app for cellphone)
Paper
Pencil (it erases...)
Periodic Table (provided on each question).

The following reaction was performed:

LiOH + KCl → LiCl + KOH
20.0 g of LiOH was reacted with an excess of KCl to produce 6.0 g LiCl.
Step 1 - Balance the equation (if you need "1" of anything, place an _ in the blank - "shift" + "hyphen").

_______ LiOH + _______ KCl → _______ LiCl + _______ KOH

Step 2 - Calculate the theoretical yield of LiCl.
Theoretical yield _______ g
Step 3 - Calculate the percent yield if 6.0 g of LiCl was actually produced.
Percent yield _______ %

The following reaction was performed:

NaCl + CaO → CaCl2 + Na2O
20.0 g of CaO is reacted with excess NaCl.
Step 1 - Balance the equation (if you need "1" of anything, place an _ in the blank - "shift" + "hyphen").

_______ NaCl + _______ CaO → _______ CaCl2 + _______ Na2O

Step 2 - Calculate the theoretical yield of Na2O.
Theoretical yield _______ g
Step 3 - Calculate the percent yield if 4.5 g Na2O was produced.
Percent yield _______% 