Industrial Stoichiometry

Industrial stoichiometry is the use of quantitative chemical calculations to determine how much reactant is needed and how much product can be produced in large-scale chemical processes.

In a laboratory, a stoichiometric calculation might involve a few grams of material. In industry, the same principles can be applied to:

• kilograms
• tonnes
• thousands of litres
• millions of moles

The scale changes, but the chemical ratios do not.

A balanced chemical equation provides the basic relationship:

reactants → products

and its coefficients tell us the mole ratios between the substances.

Industry uses these relationships to answer questions such as:

• How much raw material should be purchased?
• How much product can a factory make?
• Which reactant will run out first?
• How much reactant will remain unused?
• What percentage of the theoretical product is actually obtained?
• How can waste and cost be reduced?

Why Stoichiometry Matters in Industry

Industrial chemistry must be economically efficient.

Using too much reactant can:

• waste raw materials
• increase costs
• create additional waste
• increase separation costs
• require more storage
• increase environmental impacts

Using too little reactant may:

• reduce product output
• leave another reactant unused
• lower production efficiency

Stoichiometry helps manufacturers determine appropriate quantities before a reaction is carried out.

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From Laboratory Scale to Industrial Scale

Consider:

2H₂ + O₂ → 2H₂O

The equation means:

2 mol H₂ : 1 mol O₂ : 2 mol H₂O

This ratio applies whether the reaction involves:

• 2 mol H₂
• 2,000 mol H₂
• 2,000,000 mol H₂

The stoichiometric relationship remains:

2 : 1 : 2

This allows laboratory chemistry to be scaled up for industrial production.

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The Industrial Calculation Pathway

A useful general strategy is:

BALANCED EQUATION → MOLES → MOLE RATIO → REQUIRED QUANTITY

If masses are involved:

MASS → MOLES → MOLE RATIO → MOLES → MASS

Using:

n = m/M

and:

m = nM

If solutions are involved:

n = cV

If gases are involved, appropriate gas relationships may also be used.

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Example: Manufacturing Magnesium Oxide

Consider:

2Mg + O₂ → 2MgO

Suppose a manufacturer reacts:

240 kg Mg

with excess oxygen.

How much MgO can theoretically be produced?

Use approximate molar masses:

M(Mg) = 24 g/mol

M(MgO) = 40 g/mol

The equation shows:

2 mol Mg → 2 mol MgO

Therefore:

1 mol Mg → 1 mol MgO

The mass ratio is:

24 g Mg → 40 g MgO

Scale this directly:

240 kg Mg → 400 kg MgO

Answer

The theoretical production is:

400 kg MgO

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Mass Ratios from Balanced Equations

Balanced equations give mole ratios, but industrial quantities are often measured by mass.

Consider:

N₂ + 3H₂ → 2NH₃

Molar masses:

N₂ = 28 g/mol

H₂ = 2 g/mol

NH₃ = 17 g/mol

The balanced equation tells us:

1 mol N₂ + 3 mol H₂ → 2 mol NH₃

Convert this to masses:

28 g N₂ + 6 g H₂ → 34 g NH₃

Therefore, at any scale:

28 kg N₂ + 6 kg H₂ → 34 kg NH₃

or:

28 tonnes N₂ + 6 tonnes H₂ → 34 tonnes NH₃

assuming complete reaction and 100% yield.

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The Haber Process

One of the most important examples of industrial stoichiometry is the production of ammonia.

The Haber process uses:

N₂ + 3H₂ ⇌ 2NH₃

Ammonia is an important starting material for products including nitrogen fertilizers.

The stoichiometric ratio is:

1 mol N₂ : 3 mol H₂ : 2 mol NH₃

This means the hydrogen requirement is three times the nitrogen requirement in moles, not in mass.

Worked Example: Ammonia Production

How much ammonia can theoretically be produced from:

280 kg N₂

with excess hydrogen?

Equation:

N₂ + 3H₂ → 2NH₃

We know:

28 g N₂ → 34 g NH₃

Therefore:

280 kg N₂ → ?

Calculate:

mass NH₃ = 280 × (34 / 28)

= 340 kg

Answer

Theoretical NH₃ production = 340 kg

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Calculating Hydrogen Required

How much H₂ is required to react with:

280 kg N₂?

From:

N₂ + 3H₂ → 2NH₃

the mass relationship is:

28 g N₂ : 6 g H₂

Therefore:

280 kg N₂ : 60 kg H₂

Answer

60 kg H₂

is theoretically required.

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Industrial Scale Does Not Change the Mole Ratio

A common mistake is to think that large-scale reactions require a different calculation method.

They do not.

For:

N₂ + 3H₂ → 2NH₃

the ratio is always:

1 : 3 : 2

whether the reaction occurs in:

• a school laboratory
• a pilot plant
• a large industrial reactor

What changes are practical considerations such as:

• temperature
• pressure
• reaction rate
• catalysts
• recycling
• energy consumption
• safety
• separation of products

The stoichiometric relationship itself remains unchanged.

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Limiting Reactants in Industry

The limiting reactant is the reactant that is consumed first.

Once the limiting reactant is used up, no additional product can form unless more of that reactant is supplied.

Consider:

N₂ + 3H₂ → 2NH₃

Suppose a reactor contains:

10 mol N₂

and:

24 mol H₂

To react with 10 mol N₂, we would need:

10 × 3 = 30 mol H₂

But only:

24 mol H₂

is available.

Therefore:

H₂ is the limiting reactant.

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Determining Product from the Limiting Reactant

Continue the previous example.

We have:

24 mol H₂

From:

3H₂ → 2NH₃

calculate:

24 × (2/3) = 16 mol NH₃

Therefore:

16 mol NH₃

is the maximum theoretical amount that can form.

The amount of nitrogen available does not determine the product because hydrogen runs out first.

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Excess Reactants

An excess reactant is present in more than the amount required by the balanced equation.

Industrial processes sometimes deliberately use one reactant in excess.

Possible reasons include:

• increasing the use of a more expensive reactant
• helping drive a reaction toward products
• compensating for incomplete conversion
• improving reaction rate

However, excess reactants can also increase:

• material costs
• separation requirements
• recycling requirements
• waste

The best choice depends on the process.

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Worked Example: Limiting and Excess Reactants

Consider:

2SO₂ + O₂ → 2SO₃

Suppose a process begins with:

20 mol SO₂

and:

8 mol O₂

For 20 mol SO₂, the oxygen required is:

20 × (1/2) = 10 mol O₂

Only 8 mol O₂ is available.

Therefore:

O₂ is limiting.

Using:

1 mol O₂ → 2 mol SO₃

then:

8 mol O₂ → 16 mol SO₃

Only 16 mol SO₂ reacts.

SO₂ remaining:

20 − 16 = 4 mol

Results

Limiting reactant = O₂

SO₃ produced = 16 mol

SO₂ remaining = 4 mol

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Theoretical Yield

The theoretical yield is the maximum amount of product predicted by stoichiometry.

It assumes:

• the reaction goes completely
• no product is lost
• no unwanted reactions occur
• reactants have appropriate purity

For example, if calculations predict:

500 kg product

then:

theoretical yield = 500 kg

In real industrial processes, the actual amount obtained may be lower.

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Actual Yield

The actual yield is the amount of product actually collected or produced.

Suppose:

theoretical yield = 500 kg

but the plant obtains:

450 kg

Then:

actual yield = 450 kg

The difference can result from:

• incomplete reactions
• equilibrium limitations
• side reactions
• material losses
• separation losses
• impurities
• equipment limitations

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Percentage Yield

Percentage yield compares actual production with theoretical production.

Use:

percentage yield = (actual yield / theoretical yield) × 100

For example:

Theoretical yield:

500 kg

Actual yield:

450 kg

Then:

percentage yield = (450 / 500) × 100

= 90%

A 90% yield means the process obtained 90% of the amount predicted theoretically.

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Worked Example: Industrial Yield

A plant should theoretically produce:

800 tonnes

of a chemical.

Actual production is:

680 tonnes

Calculate the percentage yield.

percentage yield = (680 / 800) × 100

= 85%

Answer

Percentage yield = 85%

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Finding Actual Yield

Suppose a process has:

theoretical yield = 1200 kg

and:

percentage yield = 75%

Use:

actual yield = theoretical yield × percentage yield as a decimal

actual yield = 1200 × 0.75

= 900 kg

Answer

Actual yield = 900 kg

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Finding Theoretical Yield

Suppose:

actual yield = 720 kg

and:

percentage yield = 80%

Then:

0.80 = 720 / theoretical yield

Therefore:

theoretical yield = 720 / 0.80

= 900 kg

Answer

Theoretical yield = 900 kg

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Combining Stoichiometry and Percentage Yield

Industrial problems often require both calculations.

Consider:

N₂ + 3H₂ → 2NH₃

Suppose:

280 kg N₂

reacts with excess hydrogen.

We already calculated:

theoretical NH₃ yield = 340 kg

Suppose the process operates at:

80% yield

Actual ammonia production:

340 × 0.80

= 272 kg

Answer

Actual NH₃ production = 272 kg

This is much more realistic than assuming every industrial reaction achieves 100% yield.

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Working Backwards from Required Production

Industry often begins with a production target.

Suppose a factory needs to actually produce:

850 kg NH₃

and the process operates at:

85% yield.

First determine the theoretical amount required:

theoretical yield = 850 / 0.85

= 1000 kg NH₃

Now use stoichiometry to calculate the required reactants.

For:

N₂ + 3H₂ → 2NH₃

we know:

28 kg N₂ → 34 kg NH₃

Therefore:

mass N₂ = 1000 × (28 / 34)

≈ 824 kg N₂

This shows why yield must be considered when planning how much raw material to supply.

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Raw Material Purity

Industrial raw materials are not always 100% pure.

Suppose a factory needs:

500 kg pure CaCO₃

but its limestone is only:

80% CaCO₃

The required limestone mass is:

mass limestone = 500 / 0.80

= 625 kg

Answer

The factory needs:

625 kg limestone

because only 80% of the material is the desired reactant.

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Worked Example: Impure Raw Material

A process requires:

900 kg pure reactant

The available industrial material is:

75% reactant by mass

Calculate the mass of material required.

mass material = 900 / 0.75

= 1200 kg

Answer

1200 kg of raw material

must be supplied.

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Limestone and Lime Production

An important industrial reaction is the thermal decomposition of calcium carbonate:

CaCO₃ → CaO + CO₂

Calcium oxide, CaO, is commonly called quicklime.

Approximate molar masses:

CaCO₃ = 100 g/mol

CaO = 56 g/mol

CO₂ = 44 g/mol

Therefore:

100 kg CaCO₃ → 56 kg CaO + 44 kg CO₂

The same ratio applies at tonne scale:

100 tonnes CaCO₃ → 56 tonnes CaO + 44 tonnes CO₂

assuming pure material and 100% yield.

Worked Example: Quicklime Production

A plant decomposes:

5.0 tonnes CaCO₃

Calculate the theoretical mass of CaO.

From:

100 CaCO₃ → 56 CaO

calculate:

5.0 × (56 / 100)

= 2.8 tonnes

Answer

Theoretical CaO yield = 2.8 tonnes

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Including Percentage Yield

Suppose the quicklime process operates at:

90% yield

Theoretical yield:

2.8 tonnes

Actual yield:

2.8 × 0.90

= 2.52 tonnes

Answer

Actual CaO production = 2.52 tonnes

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Including Raw Material Purity

Suppose instead the plant receives:

5.0 tonnes limestone

that is only:

80% CaCO₃

First find the actual CaCO₃ mass:

5.0 × 0.80

= 4.0 tonnes CaCO₃

Now use stoichiometry:

4.0 × (56 / 100)

= 2.24 tonnes CaO theoretical

If the process has a 90% yield:

2.24 × 0.90

= 2.016 tonnes

Approximately:

2.02 tonnes CaO

This demonstrates how industrial calculations may combine:

• purity
• stoichiometry
• theoretical yield
• percentage yield

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Sulfuric Acid Production

Sulfuric acid is a major industrial chemical.

An important stage of the Contact process involves:

2SO₂ + O₂ ⇌ 2SO₃

The sulfur trioxide is then used in later stages to produce sulfuric acid.

Stoichiometric calculations help determine:

• quantities of sulfur-containing feedstocks
• oxygen requirements
• expected SO₃ production
• material flows through the plant

Industrial Efficiency

A chemically successful reaction is not automatically an efficient industrial process.

Manufacturers consider several factors.

Raw material efficiency

How much useful product is produced from the reactants?

Yield

How much of the theoretical product is actually obtained?

Energy requirements

How much energy is needed for:

• heating
• cooling
• compression
• pumping
• separation

Waste production

How much unwanted material is produced?

Recycling

Can unused reactants be returned to the reactor?

Product separation

How difficult is it to separate and purify the desired product?

Safety

Are the reactants, products, pressures, or temperatures hazardous?

Cost

Can the product be manufactured economically?

Stoichiometry provides essential information for many of these decisions.

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Recycling Unreacted Materials

Industrial reactions do not always convert all reactants during a single pass through a reactor.

Instead of discarding unused reactants, manufacturers may:

1. separate them from the product
2. return them to the reactor
3. react them again

This is called recycling.

Recycling can:

• reduce raw material consumption
• reduce waste
• improve overall process efficiency
• reduce costs

It is particularly important in processes where complete conversion during one pass is difficult.

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Stoichiometry and Waste Reduction

Suppose a reaction requires:

2 mol A : 1 mol B

If a manufacturer repeatedly supplies:

2 mol A : 5 mol B

large quantities of B may remain unused.

If B cannot be economically recovered, this creates unnecessary waste.

Stoichiometric calculations help engineers choose more appropriate feed quantities.

However, the exact industrial ratio may deliberately differ from the theoretical stoichiometric ratio when there is a practical reason.

The goal is not simply to use exact ratios blindly. The goal is to understand the chemistry well enough to optimize the entire process.

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Atom Economy

Atom economy considers how much of the reactant material becomes part of the desired product.

A simplified expression is:

atom economy = (Mr of desired product × its coefficient / total Mr of reactants using coefficients) × 100

High atom economy means a larger proportion of reactant atoms ends up in the desired product.

This is different from percentage yield.

Percentage yield asks:

How much product did we actually obtain compared with the theoretical amount?

Atom economy asks:

How much of the reactant material is designed to become useful product?

A reaction can have:

• high yield but poor atom economy
• high atom economy but poor actual yield
• both high yield and high atom economy

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Simple Atom Economy Example

Consider:

CaCO₃ → CaO + CO₂

Suppose CaO is the desired product.

Formula masses:

CaCO₃ = 100

CaO = 56

Atom economy:

(56 / 100) × 100

= 56%

The remaining mass becomes CO₂.

This tells us something important about the material efficiency of the reaction.

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Yield vs. Atom Economy

These ideas should not be confused.

Both are important when evaluating industrial chemistry.

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Process Efficiency and Sustainability

Modern chemical manufacturing increasingly aims to reduce:

• raw material consumption
• energy consumption
• hazardous materials
• unwanted by-products
• greenhouse gas emissions
• water use
• waste

Stoichiometry helps quantify material use and waste.

This makes stoichiometric calculations important not only economically but also environmentally.

Example: Comparing Two Processes

Suppose Process A uses:

1000 kg raw material

and produces:

800 kg useful product

while Process B uses:

1000 kg raw material

and produces:

500 kg useful product

If all other factors were equal, Process A would appear more materially efficient.

However, a full evaluation should also consider:

• energy requirements
• toxicity
• waste
• raw material availability
• recycling
• equipment
• reaction rate
• product purity
• environmental effects

Stoichiometry provides part of the evidence needed to make these comparisons.

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Scale-Up Calculations

Suppose a laboratory reaction uses:

5.0 g reactant A

to produce:

8.0 g product B

If the stoichiometric relationship remains the same, a manufacturer might initially predict:

500 kg A → 800 kg B

However, real scale-up must also account for:

• percentage yield
• purity
• heat transfer
• mixing
• reaction rate
• separation losses
• equipment limitations

Stoichiometry provides the theoretical foundation, while chemical engineering determines how the process performs at industrial scale.

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Multi-Step Industrial Example

Consider:

CaCO₃ → CaO + CO₂

A factory processes:

20 tonnes limestone

The limestone is:

85% CaCO₃

The process operates at:

80% yield

Calculate the actual mass of CaO produced.

Find the mass of pure CaCO₃

20 × 0.85

= 17 tonnes CaCO₃

Calculate theoretical CaO

From:

100 tonnes CaCO₃ → 56 tonnes CaO

Therefore:

17 × (56 / 100)

= 9.52 tonnes CaO

Apply percentage yield

9.52 × 0.80

= 7.616 tonnes

Answer

Approximately:

7.62 tonnes CaO

This is a realistic industrial-style problem because it combines several ideas.

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Multi-Step Example: Ammonia

A plant has:

560 kg N₂

available.

Hydrogen is supplied in excess.

Reaction:

N₂ + 3H₂ → 2NH₃

The process operates at:

75% yield

Calculate the actual mass of ammonia produced.

From the stoichiometric relationship:

28 kg N₂ → 34 kg NH₃

Therefore:

560 × (34 / 28)

= 680 kg NH₃ theoretical

Apply the yield:

680 × 0.75

= 510 kg

Answer

Actual NH₃ production = 510 kg

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Working Backwards in Manufacturing

A factory may need to produce a specific amount of product rather than simply calculate what a given amount of reactant will make.

Suppose the target is:

900 kg product

and the process has:

75% yield

The theoretical production required is:

900 / 0.75

= 1200 kg

Stoichiometry must then be used to determine how much reactant would theoretically produce 1200 kg.

This is important when planning:

• purchasing
• production schedules
• storage
• transportation
• costs

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Common Mistakes

Using Mass Ratios Directly from Coefficients

For:

N₂ + 3H₂ → 2NH₃

the coefficients give:

1 : 3 : 2 in moles

They do not mean:

1 kg N₂ : 3 kg H₂ : 2 kg NH₃

Convert between moles and mass correctly.

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Forgetting the Limiting Reactant

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

The limiting reactant determines the theoretical yield.

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Calculating Yield from the Excess Reactant

The excess reactant cannot determine the maximum product because some of it remains unused.

Use the limiting reactant.

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Confusing Actual and Theoretical Yield

Theoretical yield is calculated.

Actual yield is obtained experimentally or operationally.

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Reversing the Percentage Yield Equation

Use:

percentage yield = (actual / theoretical) × 100

not:

theoretical / actual

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Applying Yield Before Stoichiometry Incorrectly

Usually:

1. determine theoretical product from stoichiometry
2. apply percentage yield

For production targets, work backwards:

1. correct the target for percentage yield
2. determine the required reactants

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Ignoring Raw Material Purity

If limestone is:

80% CaCO₃

then:

1000 kg limestone

contains only:

800 kg CaCO₃

Do not treat the entire 1000 kg as pure reactant.

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Confusing Yield with Purity

Purity describes the fraction of the starting material that is the desired substance.

Yield describes how much product is obtained compared with the theoretical amount.

They are different corrections.

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Assuming 100% Yield in Real Processes

Theoretical calculations often begin with 100% conversion, but real manufacturing may have lower actual yields.

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Assuming More Reactant Is Always Better

Excess reactant may improve some processes, but it can also increase:

• cost
• waste
• separation requirements

Industrial chemistry involves optimization rather than simply maximizing reactant quantities.

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Key Terms

Industrial stoichiometry — The application of quantitative chemical relationships to large-scale chemical manufacturing.

Raw material — A substance used as an input in a manufacturing process.

Feedstock — Raw material supplied to an industrial chemical process.

Limiting reactant — The reactant consumed first and therefore responsible for limiting the maximum amount of product.

Excess reactant — A reactant present in more than the stoichiometrically required quantity.

Theoretical yield — The maximum amount of product predicted by stoichiometric calculations.

Actual yield — The amount of product actually obtained.

Percentage yield — The actual yield expressed as a percentage of the theoretical yield.

Purity — The fraction or percentage of a material consisting of the desired substance.

Atom economy — A measure of how much of the reactant material becomes part of the desired product.

By-product — A substance produced alongside the desired product.

Recycling — Recovering unused material and returning it to the production process.

Scale-up — Increasing a chemical process from laboratory or pilot scale to industrial production.

Process efficiency — A broad measure of how effectively a process uses materials, energy, equipment, time, and other resources.

Haber process — The industrial process used to manufacture ammonia from nitrogen and hydrogen.

Contact process — An industrial process used in the manufacture of sulfuric acid.

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Key Takeaways

• Industrial stoichiometry applies ordinary stoichiometric principles to large-scale manufacturing.
• Balanced equations provide the mole ratios used in industrial calculations.
• The same stoichiometric ratio applies whether quantities are measured in grams, kilograms, or tonnes.
• Mass calculations commonly follow:

MASS → MOLES → MOLE RATIO → MOLES → MASS

• The limiting reactant determines the maximum possible product.
• Excess reactants may remain after a reaction.
• Industries sometimes deliberately use excess reactants when there is a practical advantage.
• Theoretical yield is the maximum product predicted by stoichiometry.
• Actual yield is the amount actually produced.
• Percentage yield is:

percentage yield = (actual yield / theoretical yield) × 100

• Raw material purity must be considered when calculating industrial reactant requirements.
• Industrial problems may combine:

purity → stoichiometry → theoretical yield → percentage yield

• Recycling unused reactants can reduce costs and waste.
• Stoichiometric calculations help manufacturers plan raw material purchases and production targets.
• Better material efficiency can reduce both economic costs and environmental impacts.
• Atom economy measures how effectively reactant atoms are incorporated into the desired product.
• Percentage yield and atom economy measure different aspects of process efficiency.
• Industrial optimization also considers energy, safety, reaction rate, separation, recycling, and environmental effects.

The central industrial idea is:

USE THE RIGHT AMOUNT OF MATERIAL → MAXIMIZE USEFUL PRODUCT → MINIMIZE COST AND WASTE

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Check Your Understanding

Industrial Stoichiometry

1. Define industrial stoichiometry.

2. Explain why stoichiometric calculations are important in chemical manufacturing.

3. Why does a mole ratio remain the same when a reaction is scaled from grams to tonnes?

4. Give three problems that can result from using unnecessary excess reactants.

5. Explain how stoichiometry can reduce industrial waste.

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Mass Calculations

Use:

2Mg + O₂ → 2MgO

6. Calculate the theoretical mass of MgO produced from 120 kg Mg. Use M(Mg) = 24 g/mol and M(MgO) = 40 g/mol.

7. Calculate the theoretical mass of MgO produced from 600 kg Mg.

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Use:

N₂ + 3H₂ → 2NH₃

8. How many kilograms of NH₃ can theoretically be produced from 28 kg N₂?

9. How much H₂ is required to react completely with 140 kg N₂?

10. Calculate the theoretical NH₃ production from 560 kg N₂ with excess H₂.

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Limiting Reactants

Use:

N₂ + 3H₂ → 2NH₃

11. A reactor contains 20 mol N₂ and 45 mol H₂. Identify the limiting reactant.

12. Calculate the maximum number of moles of NH₃ that can form.

13. Determine the amount of excess reactant remaining.

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Use:

2SO₂ + O₂ → 2SO₃

14. A reactor contains 50 mol SO₂ and 20 mol O₂. Identify the limiting reactant.

15. Calculate the maximum amount of SO₃ produced.

16. Calculate the amount of excess reactant remaining.

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Percentage Yield

17. A process has a theoretical yield of 500 kg and an actual yield of 425 kg. Calculate the percentage yield.

18. A process should produce 900 tonnes but actually produces 720 tonnes. Calculate the percentage yield.

19. A reaction has a theoretical yield of 1200 kg and operates at 75% yield. Calculate the actual yield.

20. A process produces 640 kg at an 80% yield. Calculate the theoretical yield.

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Purity

21. A factory requires 800 kg of pure CaCO₃. Its limestone is 80% CaCO₃. Calculate the mass of limestone required.

22. An industrial feedstock is 75% reactant. How much feedstock is required to supply 1500 kg of pure reactant?

23. A factory receives 2000 kg of material that is 85% pure. Calculate the mass of useful reactant present.

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Lime Production

Use:

CaCO₃ → CaO + CO₂

and:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

24. Calculate the theoretical mass of CaO produced from 10 tonnes of pure CaCO₃.

25. If the process in Question 24 operates at 80% yield, calculate the actual CaO production.

26. A factory processes 10 tonnes of limestone containing 90% CaCO₃. Calculate the theoretical CaO production.

27. If the process in Question 26 operates at 75% yield, calculate the actual CaO production.

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Industrial Ammonia

Use:

N₂ + 3H₂ → 2NH₃

28. A plant uses 2800 kg N₂ with excess H₂. Calculate the theoretical NH₃ production.

29. If the process operates at 85% yield, calculate the actual NH₃ production.

30. Explain why unused reactants may be recycled in an industrial ammonia plant.

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Atom Economy and Efficiency

31. Define atom economy.

32. Explain the difference between atom economy and percentage yield.

33. For CaCO₃ → CaO + CO₂, calculate the atom economy when CaO is the desired product.

34. Can a reaction have a high percentage yield but low atom economy? Explain.

35. Give three factors other than stoichiometry that manufacturers must consider when evaluating an industrial chemical process.

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Analysis and Application

36. A plant processes 50 tonnes of raw material that is 80% pure. Calculate the mass of pure reactant available.

37. That pure reactant theoretically produces product in a 1.5:1 product-to-reactant mass ratio. Calculate the theoretical product mass from Question 36.

38. The process in Question 37 operates at 75% yield. Calculate the actual product mass.

39. Explain how limiting reactants, percentage yield, raw material purity, and recycling can all influence the economic efficiency of a chemical plant.

40. A company is considering two manufacturing methods for the same chemical. Describe the quantitative and practical factors it should compare to determine which process is more efficient, economical, and environmentally sustainable.