Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant, according to the balanced chemical equation.

It is called theoretical because it represents what should be produced under ideal conditions.

The calculation assumes that:

• the reaction goes completely to products
• no product is lost
• no unwanted side reactions occur
• the reactants are pure
• the chemical equation accurately represents the reaction

In a real experiment, the amount of product collected is often lower than the theoretical yield.

Theoretical Yield and Stoichiometry

Theoretical yield is calculated using stoichiometry.

A balanced chemical equation tells us the mole relationship between reactants and products.

For example:

2Mg + O₂ → 2MgO

This tells us:

2 mol Mg → 2 mol MgO

or:

1 mol Mg → 1 mol MgO

If we know how much magnesium reacts, we can calculate the maximum amount of magnesium oxide that could theoretically form.

The basic pathway is:

amount of reactant → moles of reactant → moles of product → theoretical yield

If the answer is required in grams:

g reactant → mol reactant → mol product → g product

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A Simple Example

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

4 mol H₂

react with sufficient oxygen.

The mole ratio is:

2 mol H₂ : 2 mol H₂O

Therefore:

4 mol H₂ → 4 mol H₂O

So the theoretical yield is:

4 mol H₂O

If we want the answer in grams:

M(H₂O) = 18 g/mol

Therefore:

m = nM

m = 4 × 18

= 72 g

Theoretical yield

72 g H₂O

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Theoretical Yield Is a Maximum

The word maximum is important.

If stoichiometry predicts:

72 g H₂O

then the theoretical yield is:

72 g

Under the assumptions of the calculation, the reaction cannot produce more product from the stated amount of limiting reactant.

In a laboratory, however, we might collect:

68 g

or:

61 g

or some other amount below the theoretical value.

The amount actually obtained is called the actual yield.

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Theoretical Yield vs. Actual Yield

Theoretical Yield

The maximum amount predicted by stoichiometry.

It is calculated.

Actual Yield

The amount actually obtained during an experiment.

It is usually measured.

For example:

Theoretical yield:

25.0 g

Actual yield:

21.3 g

The difference indicates that not all of the theoretically possible product was successfully obtained.

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Where Theoretical Yield Comes From

Theoretical yield comes from three pieces of information:

The Balanced Equation

This provides the mole ratio.

The Amount of Reactant

This tells us how much material is available.

The Molar Masses

These allow us to convert between grams and moles.

Together, these allow us to predict the maximum product.

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The Basic Calculation Method

When the mass of one reactant is given:

Balance the equation.

Convert the given mass to moles.

Use:

n = m/M

Use the mole ratio.

Convert:

mol reactant → mol product

Convert product moles to mass.

Use:

m = nM

The resulting mass is the theoretical yield.

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Worked Example: Magnesium Oxide

Consider:

2Mg + O₂ → 2MgO

What is the theoretical yield of MgO when 12.15 g Mg reacts with excess oxygen?

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Convert Mg to moles

n = 12.15 / 24.3

= 0.500 mol Mg

Use the mole ratio

From:

2Mg → 2MgO

the ratio is:

1 : 1

Therefore:

0.500 mol Mg → 0.500 mol MgO

Convert MgO to mass

m = 0.500 × 40.3

= 20.15 g

Answer

Theoretical yield = 20.15 g MgO

Approximately:

20.2 g MgO

Worked Example: Formation of Water

Consider:

2H₂ + O₂ → 2H₂O

What is the theoretical yield of water from 10.0 g H₂, assuming excess oxygen?

Use:

M(H₂) = 2.0 g/mol

M(H₂O) = 18.0 g/mol

Convert H₂ to moles

10.0 ÷ 2.0 = 5.0 mol H₂

Use the mole ratio

H₂ : H₂O = 2 : 2

Therefore:

5.0 mol H₂ → 5.0 mol H₂O

Convert to mass

5.0 × 18.0 = 90 g

Answer

Theoretical yield = 90 g H₂O

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Why the Product Can Have More Mass

In the previous example:

10 g H₂

can theoretically produce:

90 g H₂O

This does not violate conservation of mass.

Oxygen also contributes mass to the product.

The reaction requires:

5 mol H₂

and:

2.5 mol O₂

Mass of O₂:

2.5 × 32 = 80 g

Therefore:

10 g H₂ + 80 g O₂ → 90 g H₂O

Mass is conserved.

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Worked Example: Calcium Carbonate

Calcium carbonate decomposes when heated:

CaCO₃ → CaO + CO₂

What is the theoretical yield of CaO from 250 g CaCO₃?

Use:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

Convert CaCO₃ to moles

250 ÷ 100 = 2.50 mol CaCO₃

Use the mole ratio

CaCO₃ : CaO = 1 : 1

Therefore:

2.50 mol CaO

Convert to mass

2.50 × 56 = 140 g

Answer

Theoretical yield = 140 g CaO

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Predicting the Other Product

For:

CaCO₃ → CaO + CO₂

what is the theoretical yield of CO₂ from the same 250 g CaCO₃?

Use:

M(CO₂) = 44 g/mol

We already know:

250 g CaCO₃ = 2.50 mol CaCO₃

The ratio is:

1 mol CaCO₃ : 1 mol CO₂

Therefore:

2.50 mol CO₂

Mass:

2.50 × 44 = 110 g

Answer

Theoretical yield = 110 g CO₂

Check:

140 g CaO + 110 g CO₂ = 250 g

Conservation of mass is satisfied.

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Theoretical Yield with Different Coefficients

Consider:

2KClO₃ → 2KCl + 3O₂

What is the theoretical yield of oxygen from 49.0 g KClO₃?

Use:

M(KClO₃) = 122.5 g/mol

M(O₂) = 32.0 g/mol

Convert KClO₃ to moles

49.0 ÷ 122.5 = 0.400 mol KClO₃

Use the mole ratio

KClO₃ : O₂ = 2 : 3

Therefore:

0.400 × (3/2)

= 0.600 mol O₂

Convert to mass

0.600 × 32.0 = 19.2 g

Answer

Theoretical yield = 19.2 g O₂

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Theoretical Yield and Limiting Reactants

When two or more reactant quantities are given, you cannot simply choose one reactant to calculate theoretical yield.

You must first identify the limiting reactant.

The limiting reactant determines the theoretical yield because it is consumed first.

Once it runs out, no additional product can form.

The pathway becomes:

reactant amounts → identify limiting reactant → calculate product from limiting reactant → theoretical yield

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Worked Example with Two Reactants

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

10 g H₂

react with:

64 g O₂

Use:

M(H₂) = 2 g/mol

M(O₂) = 32 g/mol

Convert H₂ to moles

10 ÷ 2 = 5 mol H₂

Convert O₂ to moles

64 ÷ 32 = 2 mol O₂

The equation requires:

2 mol H₂ : 1 mol O₂

Two moles O₂ require:

4 mol H₂

We have:

5 mol H₂

Therefore:

O₂ is limiting

and H₂ is in excess.

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Calculate Theoretical Yield from the Limiting Reactant

Equation:

2H₂ + O₂ → 2H₂O

From:

2 mol O₂

we obtain:

4 mol H₂O

Mass:

4 × 18 = 72 g

Answer

Theoretical yield = 72 g H₂O

We must use the limiting reactant because it determines the maximum amount of product.

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What If We Used the Wrong Reactant?

Suppose we incorrectly calculated the product from all 5 mol H₂.

The 1 : 1 ratio between H₂ and H₂O would predict:

5 mol H₂O

or:

90 g H₂O

But only enough oxygen exists to produce:

72 g H₂O

Therefore:

90 g is impossible under the stated conditions.

This demonstrates why identifying the limiting reactant is essential.

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Worked Example: Iron Oxide

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose:

112 g Fe

react with:

64 g O₂

Use:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

Convert reactants to moles

Fe:

112 ÷ 56 = 2 mol Fe

O₂:

64 ÷ 32 = 2 mol O₂

Identify the limiting reactant

Two moles Fe require:

2 × (3/4) = 1.5 mol O₂

We have:

2 mol O₂

Therefore:

Fe is limiting

and O₂ is in excess.

Calculate product

Ratio:

4 mol Fe : 2 mol Fe₂O₃

Therefore:

2 mol Fe → 1 mol Fe₂O₃

Mass:

1 × 160 = 160 g

Answer

Theoretical yield = 160 g Fe₂O₃

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Worked Example: Aluminum Chloride

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose:

27.0 g Al

react with:

71.0 g Cl₂

Use:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

M(AlCl₃) = 133.5 g/mol

Convert to moles

Al:

27.0 ÷ 27.0 = 1.00 mol

Cl₂:

71.0 ÷ 71.0 = 1.00 mol

The equation requires:

2 mol Al : 3 mol Cl₂

Divide by coefficients:

Al:

1.00 ÷ 2 = 0.500

Cl₂:

1.00 ÷ 3 = 0.333

Therefore:

Cl₂ is limiting

Calculate theoretical yield

Ratio:

3 mol Cl₂ : 2 mol AlCl₃

Therefore:

1.00 mol Cl₂ × (2/3)

= 0.667 mol AlCl₃

Mass:

0.667 × 133.5 ≈ 89.0 g

Answer

Theoretical yield ≈ 89.0 g AlCl₃

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Theoretical Yield and Actual Experiments

Real chemical reactions rarely behave perfectly.

Suppose stoichiometry predicts:

50.0 g product

but the experiment produces:

43.2 g product

Then:

theoretical yield = 50.0 g

actual yield = 43.2 g

The theoretical calculation has not necessarily been wrong.

Instead, practical factors may have prevented all of the theoretical product from being collected.

Why Theoretical Yield Is Often Not Achieved

There are many possible reasons.

The Reaction May Not Go to Completion

Some reactants may remain unreacted.

If not all of the limiting reactant becomes product, the actual yield will be lower.

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Product May Be Lost During Transfer

Some product may remain:

• inside a beaker
• on a stirring rod
• on filter paper
• inside a funnel
• in another piece of apparatus

Small losses can occur every time material is transferred.

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Product May Be Lost During Filtration

A precipitate may:

• pass through the filter
• remain dissolved
• stick to glassware
• be spilled

This reduces the amount collected.

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Side Reactions May Occur

Reactants may undergo unwanted reactions that form other products.

This means some reactant is used without forming the desired product.

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Reactants May Contain Impurities

Suppose a sample has a mass of:

10.0 g

but only:

8.5 g

is actually the desired reactant.

Using the full 10.0 g in a theoretical calculation would overestimate how much product can form.

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Some Product May Remain Dissolved

When a solid product forms in solution, some may remain dissolved rather than being collected.

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Gas May Escape

If the desired product is a gas, some may escape before it is collected or measured.

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Reversible Reactions May Not Go to Completion

Some reactions reach equilibrium rather than converting all reactants into products.

This can reduce the amount of desired product.

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Experimental Error and Theoretical Yield

Measurement uncertainty can also affect the comparison between theoretical and actual yield.

Possible sources include:

• balance uncertainty
• inaccurate volume measurements
• incomplete drying
• loss during heating
• contamination
• incomplete collection

The theoretical yield represents an ideal prediction, while the actual yield reflects the real experimental process.

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Can Actual Yield Be Greater Than Theoretical Yield?

In a correctly performed and interpreted experiment, the actual amount of pure desired product should not exceed the theoretical yield calculated from the true limiting reactant.

However, an experiment may appear to produce more than 100% of the theoretical yield.

For example:

Theoretical yield:

10.0 g

Measured product:

11.2 g

Possible explanations include:

• the product was wet
• impurities were present
• unreacted reactant remained with the product
• another substance contaminated the sample
• the theoretical calculation was incorrect
• the limiting reactant was identified incorrectly

So a measured mass greater than theoretical yield is usually evidence that something needs to be investigated.

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Theoretical Yield and Conservation of Mass

Theoretical yield must be consistent with conservation of mass.

Consider:

2Mg + O₂ → 2MgO

Suppose:

24.3 g Mg

react completely.

This requires:

16.0 g O₂

Total reacting mass:

24.3 + 16.0 = 40.3 g

Therefore:

theoretical yield = 40.3 g MgO

The predicted product mass exactly matches the total mass of reactants consumed.

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Theoretical Yield and Excess Reactants

An excess reactant does not increase the theoretical yield once the limiting reactant has been completely consumed.

Suppose:

2H₂ + O₂ → 2H₂O

You have:

4 mol H₂

and:

10 mol O₂

Only:

2 mol O₂

are needed to react with the 4 mol H₂.

Adding even more oxygen cannot produce more water because the hydrogen has already been completely consumed.

Therefore:

H₂ is limiting

and:

theoretical yield = 4 mol H₂O

The remaining oxygen is simply excess reactant.

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Increasing Theoretical Yield

To increase the theoretical yield, you generally need to increase the amount of the limiting reactant.

Adding more excess reactant will not increase the maximum product.

For example:

2H₂ + O₂ → 2H₂O

Suppose:

2 mol H₂ + 10 mol O₂

Hydrogen is limiting.

Adding another 5 mol O₂ changes nothing.

But increasing H₂ can increase the theoretical yield.

This is an important idea in industrial chemistry.

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Theoretical Yield in Manufacturing

Chemical manufacturers use theoretical yield calculations to predict how much product should be possible from their raw materials.

These calculations help determine:

• how much reactant to purchase
• expected production levels
• production costs
• equipment requirements
• waste quantities
• process efficiency

A factory may compare its actual production with theoretical yield to determine how effectively the process is operating.

Theoretical Yield in Pharmaceutical Chemistry

Pharmaceutical manufacturing requires careful control of chemical quantities.

Suppose a synthesis theoretically produces:

100 kg

of a pharmaceutical compound.

If the process consistently produces only:

65 kg

chemists may investigate:

• incomplete reactions
• side reactions
• purification losses
• inefficient separation
• decomposition of the product

Improving the process can increase actual production without necessarily increasing the amount of starting material.

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Theoretical Yield and Green Chemistry

Higher actual yields can often mean that fewer resources are wasted.

Low yields may result in:

• wasted reactants
• additional solvent use
• greater energy consumption
• more waste requiring disposal
• higher production costs

For this reason, improving reaction efficiency is an important goal of green chemistry.

However, yield is only one measure of sustainability. A high-yield reaction can still create significant waste or require hazardous materials.

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A Complete Problem-Solving Strategy

For a theoretical-yield problem:

Balance the chemical equation.

Never perform stoichiometry using an unbalanced equation.

Determine what information is given.

Are you given:

• moles?
• mass?
• quantities of two reactants?

Convert to moles if necessary.

Use:

n = m/M

Identify the limiting reactant if necessary.

If quantities of multiple reactants are provided, determine which runs out first.

Use the mole ratio.

Convert:

mol limiting reactant → mol product

Convert product to the requested unit.

For mass:

m = nM

State the theoretical yield clearly.

Include:

• numerical value
• unit
• substance

For example:

The theoretical yield is 35.6 g CaO.

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One-Line Stoichiometric Method

A theoretical-yield calculation can also be written as one continuous calculation.

Consider:

2Mg + O₂ → 2MgO

Starting with:

12.15 g Mg

Calculation:

12.15 g Mg × (1 mol Mg / 24.3 g Mg) × (2 mol MgO / 2 mol Mg) × (40.3 g MgO / 1 mol MgO)

= 20.15 g MgO

The units cancel:

g Mg → mol Mg → mol MgO → g MgO

Therefore:

theoretical yield = 20.15 g MgO

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Worked Example: Methane Combustion

Consider:

CH₄ + 2O₂ → CO₂ + 2H₂O

What is the theoretical yield of CO₂ from 32 g CH₄, assuming excess oxygen?

Use:

M(CH₄) = 16 g/mol

M(CO₂) = 44 g/mol

Convert CH₄ to moles

32 ÷ 16 = 2 mol CH₄

Use the mole ratio

CH₄ : CO₂ = 1 : 1

Therefore:

2 mol CO₂

Convert to mass

2 × 44 = 88 g

Answer

Theoretical yield = 88 g CO₂

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Worked Example: Sodium Chloride

Consider:

2Na + Cl₂ → 2NaCl

Suppose:

46 g Na

react with excess chlorine.

Use:

M(Na) = 23 g/mol

M(NaCl) = 58.5 g/mol

Convert Na to moles

46 ÷ 23 = 2 mol Na

Use the ratio

2 mol Na → 2 mol NaCl

Therefore:

2 mol NaCl

Convert to mass

2 × 58.5 = 117 g

Answer

Theoretical yield = 117 g NaCl

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More Challenging Example

Consider:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Suppose:

22 g C₃H₈

react with excess oxygen.

Use:

M(C₃H₈) = 44 g/mol

M(CO₂) = 44 g/mol

Find the theoretical yield of CO₂.

Convert propane to moles

22 ÷ 44 = 0.500 mol C₃H₈

Use the mole ratio

C₃H₈ : CO₂ = 1 : 3

Therefore:

0.500 × 3 = 1.50 mol CO₂

Convert to mass

1.50 × 44 = 66 g

Answer

Theoretical yield = 66 g CO₂

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Common Mistakes

Forgetting to Balance the Equation

The mole ratio comes from the balanced equation.

An incorrect equation produces an incorrect theoretical yield.

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Using Grams Directly with the Coefficients

Coefficients represent mole ratios, not mass ratios.

Use:

grams → moles → mole ratio → grams

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Ignoring the Limiting Reactant

If two reactant amounts are given, theoretical yield must be based on the limiting reactant.

Using the excess reactant will overestimate the yield.

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Choosing the Smaller Mass as Limiting

The reactant with fewer grams is not automatically limiting.

Convert to moles and compare using the balanced equation.

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Choosing the Smaller Number of Moles as Limiting

The balanced equation may require unequal numbers of moles.

Always consider the coefficients.

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Confusing Theoretical and Actual Yield

Theoretical yield is calculated.

Actual yield is measured experimentally.

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Assuming Theoretical Yield Is Always Obtained

Theoretical yield represents ideal conditions.

Real experiments usually involve some loss or inefficiency.

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Thinking a Product Cannot Have More Mass Than One Reactant

Other reactants also contribute mass to the product.

Compare total reacting mass, not just one reactant.

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Accepting More Than 100% Without Investigation

An apparent yield greater than the theoretical amount usually suggests:

• contamination
• incomplete drying
• measurement error
• incorrect calculations

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Rounding Too Early

Keep several digits during intermediate calculations and round the final result appropriately.

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Key Terms

Theoretical yield — The maximum amount of product predicted by stoichiometry from the available limiting reactant.

Actual yield — The amount of product actually obtained experimentally.

Stoichiometry — The quantitative relationship between reactants and products in chemical reactions.

Limiting reactant — The reactant consumed first and therefore responsible for determining theoretical yield.

Excess reactant — A reactant present in more than the stoichiometric amount required.

Mole ratio — The ratio between substances given by coefficients in a balanced equation.

Molar mass — The mass of one mole of a substance, expressed in g/mol.

Balanced equation — A chemical equation containing equal numbers of each type of atom on both sides.

Maximum yield — Another way of describing the greatest quantity of product theoretically possible.

Side reaction — An unwanted reaction that consumes reactants or products and forms substances other than the desired product.

Reaction completion — The extent to which the available limiting reactant has been converted into products.

Product loss — Desired product that forms but is not successfully collected or measured.

Experimental error — Measurement or procedural uncertainty that affects experimental results.

Purity — The proportion of a sample consisting of the desired substance rather than impurities.

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Key Takeaways

• Theoretical yield is the maximum amount of product predicted by stoichiometry.
• It is calculated from a balanced chemical equation.
• The calculation assumes ideal reaction conditions.
• When one reactant quantity is given and other reactants are in excess, use the given reactant to calculate theoretical yield.
• When multiple reactant quantities are given, first identify the limiting reactant.
• The limiting reactant determines theoretical yield.
• Excess reactant cannot produce additional product after the limiting reactant has been consumed.
• The basic mass pathway is:

g reactant → mol reactant → mol product → g product

• Molar mass converts between grams and moles.
• Balanced-equation coefficients provide the mole ratio.
• Actual yield is the amount obtained experimentally.
• Actual yield is often lower than theoretical yield.
• Product can be lost through transfers, filtration, purification, heating, or other procedures.
• Side reactions, impurities, incomplete reactions, and equilibrium can also reduce actual yield.
• An apparent actual yield greater than theoretical yield should be investigated.
• Increasing the excess reactant alone does not increase theoretical yield.
• Increasing the limiting reactant can increase theoretical yield.
• Theoretical-yield calculations are important in laboratory chemistry, manufacturing, pharmaceuticals, environmental chemistry, and process design.

The main strategy is:

BALANCE → MOLES → IDENTIFY LIMITING REACTANT → MOLE RATIO → PRODUCT → THEORETICAL YIELD

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Check Your Understanding

Use:

2Mg + O₂ → 2MgO

with:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

1. Define theoretical yield.

2. Calculate the theoretical yield of MgO from 24.3 g Mg, assuming excess oxygen.

3. Calculate the theoretical yield of MgO from 48.6 g Mg.

4. Calculate the theoretical yield of MgO from 6.075 g Mg.

5. Explain why the mass of MgO can be greater than the initial mass of Mg.

Use:

CaCO₃ → CaO + CO₂

with:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

M(CO₂) = 44 g/mol

6. Calculate the theoretical yield of CaO from 100 g CaCO₃.

7. Calculate the theoretical yield of CO₂ from 100 g CaCO₃.

8. Calculate both theoretical yields from 350 g CaCO₃.

9. Add the masses of the two products from Question 8. What do you notice?

10. Explain how your answer demonstrates conservation of mass.

Limiting Reactant Problems

Use:

2H₂ + O₂ → 2H₂O

with:

M(H₂) = 2 g/mol

M(O₂) = 32 g/mol

M(H₂O) = 18 g/mol

11. If 8 g H₂ react with 32 g O₂, identify the limiting reactant.

12. Calculate the theoretical yield of H₂O.

13. Calculate the mass of excess reactant remaining.

14. If 4 g H₂ react with 64 g O₂, calculate the theoretical yield.

15. Explain why adding even more excess reactant would not increase the theoretical yield.

Use:

4Fe + 3O₂ → 2Fe₂O₃

with:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

16. If 112 g Fe react with 96 g O₂, identify the limiting reactant.

17. Calculate the theoretical yield of Fe₂O₃.

18. Calculate the mass of excess reactant remaining.

19. If 224 g Fe react with 96 g O₂, calculate the theoretical yield of Fe₂O₃.

20. Determine whether any reactant remains after Question 19.

More Challenging Problems

Use:

2Al + 3Cl₂ → 2AlCl₃

with:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

M(AlCl₃) = 133.5 g/mol

21. If 54.0 g Al react with 213 g Cl₂, calculate the theoretical yield of AlCl₃.

22. If 54.0 g Al react with 142 g Cl₂, identify the limiting reactant.

23. Calculate the theoretical yield for Question 22.

24. Calculate the mass of excess reactant remaining.

25. Explain why the limiting reactant, rather than the excess reactant, must be used to determine theoretical yield.

Application and Reasoning

26. Explain the difference between theoretical yield and actual yield.

27. Give three reasons why actual yield may be lower than theoretical yield.

28. Explain how product lost during filtration affects actual yield.

29. Explain how a side reaction can reduce the amount of desired product.

30. Explain how impure reactants can affect a theoretical-yield calculation.

31. A calculation predicts a theoretical yield of 50.0 g. An experiment produces 43.0 g. Which value is the theoretical yield and which is the actual yield?

32. A reaction has a theoretical yield of 20.0 g, but a student measures 21.8 g of product. Give two possible explanations.

33. Explain why an apparent actual yield greater than theoretical yield should be investigated.

34. A manufacturer doubles the amount of an excess reactant but keeps the limiting reactant unchanged. What happens to the theoretical yield? Explain.

35. How could a manufacturer increase the theoretical yield of a process?

36. Explain why theoretical-yield calculations are useful before conducting a laboratory experiment.

37. Explain why theoretical-yield calculations are important in industrial chemistry.

38. Describe the complete procedure for calculating theoretical yield when the mass of one reactant is given and all other reactants are in excess.

39. Describe how the procedure changes when the masses of two reactants are given.

40. Explain why theoretical yield represents an ideal maximum rather than a guarantee of how much product will actually be collected.