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

Poslední aktualizace about 1 month ago

5 Dotazy

Poznámka od autora:

Instrukce

Untitled Section

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

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.

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.

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Practice Questions

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

Poslední aktualizace about 1 month ago

5 Dotazy

Poznámka od autora:

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.

Instrukce

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

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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

null

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

null

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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

Set-up the equation (3 parts)

Mass of reactant to moles of reactant (need the molar mass of reactant).

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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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

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).

null

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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

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).

null

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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

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.

null

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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

Theoretical Yield = 0.507 g Cu

Step 3 - Percent Yield

Actual Yield = 0.392 g Cu

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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

null

Step 2 - Theoretical Yield

Need molar masses of CuSO₄ and Cu
Use periodic table.

null

Theoretical Yield = 0.507 g Cu

Step 3 - Percent Yield

Actual Yield = 0.392 g Cu

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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.

null

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.

null

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.

null

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.

null

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.

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

Set-up the equation (3 parts)

Mass of reactant to moles of reactant (need the molar mass of reactant.)

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

Set-up the equation (3 parts)

Mass of reactant to moles of reactant.
Moles of reactant to moles of product (1 mol CCl₄ → 1 mol CF₂Cl₂).

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

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 (use molar mass of product).

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

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.

null

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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

Theoretical Yield = 25.86 g CF₂Cl₂

Step 3 - Percent Yield

Actual Yield = 12.5 g CF₂Cl₂

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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.

null

Step 2 - Theoretical Yield

Need molar mass of CCl₄ and CF₂Cl₂
Use periodic table.

null

Theoretical Yield = 25.86 g CF₂Cl₂

Step 3 - Percent Yield

Actual Yield = 12.5 g CF₂Cl₂

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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.

null

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.

null

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.

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

Mass of reactant to moles of reactant (use molar mass of Si).

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

Mass of reactant to moles of reactant.
Moles of reactant to moles of product (3 mol Si → 1 mol Si₃N₂).

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

Mass of reactant to moles of reactant.
Moles of reactant to moles of product.
Moles of product to mass of product (use molar mass of product).

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

Mass of reactant to moles of reactant.

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

Mass of reactant to moles of reactant.
Moles of reactant to moles of product (1 mol N₂ → 1 mol Si₃N₂).

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

Mass of reactant to moles of reactant.
Moles of reactant to moles of product (1 mol N₂ → 1 mol Si₃N₂).
Moles of product to mass of product (use the molar mass of Si₃N₂).

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Set-up two reactions (for both reactants) - 3 parts.

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

null

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

null

Step 3 - Limiting Reagent

Compare the theoretical yields.
The lowest one is 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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

null

Step 3 - Limiting Reagent

Compare the theoretical yields.
The lowest one is the limiting reagent.

Si → 2.70 g Si₃N₂

N2 → 6.01 g Si₃N₂

Si will be consumed first, since it will produce the least amount of product.

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.

null

Step 2 - Theoretical Yield.

Molar masses

null

Reaction 1 - for silicon (Si):

null

Reaction 2 - for nitrogen (N₂):

null

Step 3 - Limiting Reagent

Compare the theoretical yields.
The lowest one is the limiting reagent.

Si → 2.70 g Si₃N₂

N2 → 6.01 g Si₃N₂

Si will be consumed first, since it will produce the least amount of product.

Step 4 - Percent Yield

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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 %

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The following reaction was performed:

C₃H₈ + O₂ → CO₂ + H₂O

5.0 g C₃H₈ is reacted with excess O₂

Step 1 - Balance the equation (if you need "1" of anything, place an _ in the blank - "shift" + "hyphen").

C₃H₈ + O₂ → CO₂ + H₂O

Step 2 - Calculate the theoretical yield of H₂O.

Theoretical yield g

Step 3 - Calculate the percent yield if 7.9 g of H₂O was actually produced.

Percent yield %

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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 %

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The following reaction was performed:

Sn₃(PO₄)₄ + Na₂CO₃ → Sn(CO₃)₂ + Na₃PO₄

36 grams Sn₃(PO₄)₄is reacted with an excess ofNa₂CO₃

Step 1 - Balance the equation (if you need "1" of anything, place an _ in the blank - "shift" + "hyphen").

Sn₃(PO₄)₄ + Na₂CO₃ → Sn(CO₃)₂ + Na₃PO₄

Step 2 - Calculate the theoretical yield of Sn(CO₃)₂

Theoretical Yield of Sn(CO₃)₂ g

Step 3 - Percent yield, if 29.8 g Sn(CO₃)₂are actually produced.

Percent yield %

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The following reaction was performed:

H₂ + N₂ → NH₃

5.0 grams N₂ and 5.0 grams H₂were reacted to form 3.25 g of ammonia.

Step 1 - Balance the equation (if you need "1" of an anything, place an _ in the blank - "shift" + "hyphen").

H₂ + N₂ → NH₃

Step 2 - Determine the mass of the product produced by each reactant.

For H2 g
For N2 g

Step 3 - Compare the amounts calculated in step 2 to identify the limiting reagent.

Limiting Reagent

Step 4 - The lowest quantity calculated in step 2 will be the theoretical yield.

Theoretical Yield g

Step 5 - Calculate the percent yield if the actual product produced was measured to be 3.25g.

Percent yield %