Stoichiometric Calculations

Stoichiometry is the quantitative study of the amounts of reactants and products involved in chemical reactions.

A balanced chemical equation acts like a chemical recipe. It tells us the relative amounts of substances that react and form.

For example:

2H₂ + O₂ → 2H₂O

This tells us:

2 mol H₂ + 1 mol O₂ → 2 mol H₂O

If we know the amount of one substance, we can use the balanced equation to calculate the amount of another.

This is the central idea of stoichiometry:

known amount → balanced equation → unknown amount

The Stoichiometric Relationship

Consider:

N₂ + 3H₂ → 2NH₃

The coefficients tell us:

1 mol N₂ : 3 mol H₂ : 2 mol NH₃

From this equation we can determine many relationships.

For every:

1 mol N₂

we need:

3 mol H₂

and theoretically produce:

2 mol NH₃

If the amount of one substance changes, the amounts of the others change proportionally.

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The Mole Is the Bridge

The most important idea in stoichiometric calculations is that moles connect substances in a chemical equation.

A balanced equation directly relates:

moles ↔ moles

It does not directly relate grams to grams.

Therefore, if a question gives a quantity other than moles, we normally convert it into moles first.

The general pathway is:

given quantity → moles of given substance → mole ratio → moles of wanted substance → wanted quantity

This pathway is the foundation of most stoichiometry problems.

The Four Main Steps

A reliable approach is:

Balance the equation

Make sure the chemical equation is balanced.

Convert the given quantity to moles

If the question already gives moles, this step is unnecessary.

Use the mole ratio

Use coefficients from the balanced equation to convert between substances.

Convert to the requested quantity

If the answer is required in moles, stop.

If it is required in another quantity, perform the necessary conversion.

A useful summary is:

BALANCE → CONVERT TO MOLES → USE MOLE RATIO → CONVERT TO ANSWER

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Mole-to-Mole Calculations

The simplest stoichiometric problems give one quantity in moles and ask for another quantity in moles.

Consider:

2H₂ + O₂ → 2H₂O

Suppose 5 mol H₂ react with sufficient oxygen.

How many moles of water can form?

Identify what is given

5 mol H₂

Identify what is wanted

mol H₂O

Find the mole ratio

From the equation:

2 mol H₂ : 2 mol H₂O

Calculate

5 mol H₂ × (2 mol H₂O / 2 mol H₂)

= 5 mol H₂O

Answer

5 mol H₂O

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A Shortcut Formula

For mole-to-mole calculations:

moles wanted = moles given × (coefficient wanted / coefficient given)

For example:

N₂ + 3H₂ → 2NH₃

If we have 9 mol H₂:

moles NH₃ = 9 × (2/3)

= 6 mol NH₃

This formula is useful, but understanding the mole-ratio method is more important than memorizing the formula.

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Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

How many moles of NH₃ can form from 7.5 mol N₂?

Ratio:

1 mol N₂ : 2 mol NH₃

Calculation:

7.5 mol N₂ × (2 mol NH₃ / 1 mol N₂)

= 15 mol NH₃

Answer

15 mol NH₃

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Worked Example: Finding a Reactant

Using:

N₂ + 3H₂ → 2NH₃

How many moles of H₂ are required to produce 12 mol NH₃?

Ratio:

3 mol H₂ : 2 mol NH₃

Calculation:

12 mol NH₃ × (3 mol H₂ / 2 mol NH₃)

= 18 mol H₂

Answer

18 mol H₂

Notice that stoichiometry can be used in either direction:

reactant → product

or:

product → reactant

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Worked Example: Combustion

Methane burns according to:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many moles of oxygen are required to burn 4.5 mol CH₄?

Ratio:

1 mol CH₄ : 2 mol O₂

Calculation:

4.5 mol CH₄ × (2 mol O₂ / 1 mol CH₄)

= 9 mol O₂

Answer

9 mol O₂

Multi-Step Stoichiometry

More advanced questions may require several conversions.

For example:

mass → moles → mole ratio → moles → mass

This is one of the most common stoichiometric pathways.

The key idea is that the mole ratio always connects the two different substances.

For example:

grams A → mol A → mol B → grams B

Notice where the chemical identity changes:

mol A → mol B

That is the mole-ratio step.

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Mass and Moles

To convert between mass and moles:

n = m / M

where:

• n = amount in moles (mol)
• m = mass (g)
• M = molar mass (g/mol)

Rearranging:

m = nM

Therefore:

mass → moles

use:

n = m / M

and:

moles → mass

use:

m = nM

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Worked Example: Mass to Moles to Moles

Consider:

2Mg + O₂ → 2MgO

Suppose 12.0 g Mg reacts with sufficient oxygen.

How many moles of MgO can form?

Use:

M(Mg) ≈ 24.3 g/mol

Convert Mg to moles

n = m / M

n = 12.0 / 24.3

n ≈ 0.494 mol Mg

Use the mole ratio

From:

2Mg + O₂ → 2MgO

Mg : MgO is:

2 : 2

or:

1 : 1

Therefore:

0.494 mol Mg × (2 mol MgO / 2 mol Mg)

= 0.494 mol MgO

Answer

0.494 mol MgO

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Worked Example: Mass to Mass

Now suppose we want the mass of MgO produced.

Equation:

2Mg + O₂ → 2MgO

Given:

12.0 g Mg

Molar masses:

Mg ≈ 24.3 g/mol

MgO ≈ 40.3 g/mol

Convert Mg to moles

12.0 g ÷ 24.3 g/mol = 0.494 mol Mg

Apply the mole ratio

Mg : MgO = 1 : 1

Therefore:

0.494 mol MgO

Convert MgO to mass

m = nM

m = 0.494 × 40.3

m ≈ 19.9 g

Answer

19.9 g MgO

The complete pathway was:

12.0 g Mg → 0.494 mol Mg → 0.494 mol MgO → 19.9 g MgO

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Dimensional Analysis

The same calculation can be written as one continuous calculation:

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

Units cancel:

g Mg → mol Mg → mol MgO → g MgO

leaving:

19.9 g MgO

This method is called dimensional analysis.

It is extremely useful because the units show whether the calculation has been set up correctly.

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The Stoichiometry Road Map

Many problems can be understood using this structure:

mass of A

↓

moles of A

↓

MOLE RATIO

↓

moles of B

↓

mass of B

The middle step is always based on the balanced equation.

Worked Example: Producing Water

Consider:

2H₂ + O₂ → 2H₂O

How many grams of water can theoretically form from 8.0 g H₂, assuming sufficient oxygen?

Use approximate molar masses:

H₂ = 2.0 g/mol

H₂O = 18.0 g/mol

Convert H₂ to moles

8.0 g ÷ 2.0 g/mol = 4.0 mol H₂

Use the mole ratio

H₂ : H₂O = 2 : 2 = 1 : 1

Therefore:

4.0 mol H₂O

Convert water to mass

4.0 × 18.0 = 72 g

Answer

72 g H₂O

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Worked Example: Oxygen Required

Using:

2H₂ + O₂ → 2H₂O

How many grams of oxygen are needed to react completely with 6.0 mol H₂?

Use the mole ratio

H₂ : O₂ = 2 : 1

6.0 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 3.0 mol O₂

Molar mass:

O₂ = 32.0 g/mol

Convert to mass

m = nM

m = 3.0 × 32.0

= 96 g

Answer

96 g O₂

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Worked Example: Propane Combustion

Propane burns according to:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

How many moles of CO₂ form when 2.5 mol C₃H₈ burns completely?

Mole ratio

C₃H₈ : CO₂ = 1 : 3

Calculate

2.5 mol C₃H₈ × (3 mol CO₂ / 1 mol C₃H₈)

= 7.5 mol CO₂

Answer

7.5 mol CO₂

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Multi-Step Propane Problem

How many grams of CO₂ can form when 44 g C₃H₈ burns completely?

Equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Approximate molar masses:

C₃H₈ = 44 g/mol

CO₂ = 44 g/mol

Convert propane to moles

44 g ÷ 44 g/mol = 1 mol C₃H₈

Use the mole ratio

1 mol C₃H₈ : 3 mol CO₂

Therefore:

3 mol CO₂

Convert to mass

3 mol × 44 g/mol = 132 g

Answer

132 g CO₂

The pathway was:

44 g C₃H₈ → 1 mol C₃H₈ → 3 mol CO₂ → 132 g CO₂

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Stoichiometry with Decomposition Reactions

Stoichiometry works with any correctly balanced reaction.

Consider the decomposition of calcium carbonate:

CaCO₃ → CaO + CO₂

The ratio is:

1 : 1 : 1

If 2.5 mol CaCO₃ decomposes completely:

2.5 mol CaCO₃ → 2.5 mol CaO + 2.5 mol CO₂

Worked Example: Calcium Carbonate

How many grams of CO₂ can form from 100 g CaCO₃?

Approximate molar masses:

CaCO₃ = 100 g/mol

CO₂ = 44 g/mol

Convert CaCO₃ to moles

100 g ÷ 100 g/mol = 1 mol CaCO₃

Apply the mole ratio

CaCO₃ : CO₂ = 1 : 1

Therefore:

1 mol CO₂

Convert to mass

1 × 44 = 44 g

Answer

44 g CO₂

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Stoichiometry with Synthesis Reactions

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose 8 mol Fe reacts with sufficient oxygen.

How many moles of Fe₂O₃ can form?

Ratio:

4 mol Fe : 2 mol Fe₂O₃

Calculation:

8 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 4 mol Fe₂O₃

Answer

4 mol Fe₂O₃

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A More Complex Mass Calculation

Consider:

4Fe + 3O₂ → 2Fe₂O₃

How many grams of Fe₂O₃ can theoretically form from 112 g Fe?

Use approximate molar masses:

Fe = 56 g/mol

Fe₂O₃ = 160 g/mol

Convert Fe to moles

112 ÷ 56 = 2 mol Fe

Use the mole ratio

2 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 1 mol Fe₂O₃

Convert to mass

1 × 160 = 160 g

Answer

160 g Fe₂O₃

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Why the Product Can Have More Mass Than One Reactant

In the previous example:

112 g Fe → 160 g Fe₂O₃

Does this violate conservation of mass?

No.

The iron combines with oxygen from O₂.

The additional mass comes from oxygen.

The complete reaction conserves mass:

mass of all reactants = mass of all products

This is an important point when interpreting stoichiometric calculations.

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Starting with the Product

Stoichiometry does not always move from reactant to product.

Suppose:

2KClO₃ → 2KCl + 3O₂

How many moles of KClO₃ are required to produce 9 mol O₂?

Ratio:

2 mol KClO₃ : 3 mol O₂

Calculation:

9 mol O₂ × (2 mol KClO₃ / 3 mol O₂)

= 6 mol KClO₃

Answer

6 mol KClO₃

The calculation can move backward through the equation.

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Multi-Step Problems with Several Substances

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose we want to determine how much chlorine is needed to produce 26.7 g AlCl₃.

Approximate molar mass:

AlCl₃ = 133.5 g/mol

Convert AlCl₃ to moles

26.7 ÷ 133.5 = 0.200 mol AlCl₃

Use the mole ratio

Cl₂ : AlCl₃ = 3 : 2

0.200 mol AlCl₃ × (3 mol Cl₂ / 2 mol AlCl₃)

= 0.300 mol Cl₂

If the question asks for moles, stop here.

Answer

0.300 mol Cl₂

If mass were requested, we would continue by multiplying by the molar mass of Cl₂.

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Deciding Which Conversion to Use

Ask:

What unit do I have?

and:

What unit do I need?

If you have grams:

grams → moles

If you have moles:

you may be ready for the mole ratio.

If the answer requires grams:

moles → grams

This prevents unnecessary calculations.

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The Mole Ratio Is the Chemical Bridge

Suppose substances A and B participate in a reaction.

You cannot normally jump directly from:

grams A → grams B

Instead:

grams A → mol A → mol B → grams B

The conversion:

mol A → mol B

comes from the balanced equation.

This is the chemical bridge between the two substances.

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Why Coefficients Matter

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose you have 6 mol Al.

A common mistake would be to assume you need 6 mol Cl₂.

But the equation says:

2 mol Al : 3 mol Cl₂

Therefore:

6 mol Al × (3 mol Cl₂ / 2 mol Al)

= 9 mol Cl₂

Stoichiometry depends on the actual coefficients, not simply on the number of substances present.

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Multi-Step Calculation Strategy

When facing a longer problem, write this at the top of your page:

GIVEN → mol GIVEN → mol WANTED → WANTED

Then fill in the quantities.

For example:

24.3 g Mg → mol Mg → mol MgO → g MgO

This gives you a roadmap before you calculate anything.

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Real-World Connection: Chemical Manufacturing

Industrial chemists use stoichiometry to calculate how much raw material is required to manufacture products.

For example, ammonia is produced using:

N₂ + 3H₂ ⇌ 2NH₃

The balanced equation gives a mole ratio of:

1 mol N₂ : 3 mol H₂ : 2 mol NH₃

Manufacturers need to know:

• how much nitrogen is required
• how much hydrogen is required
• how much ammonia could theoretically form
• how efficiently raw materials are being used

Real-World Connection: Pharmaceuticals

Pharmaceutical manufacturing also requires careful control of quantities.

Using incorrect proportions can:

• waste expensive reactants
• reduce product formation
• produce unwanted by-products
• make purification more difficult

Stoichiometry allows chemists to predict how much starting material is required to produce a desired amount of product.

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Real-World Connection: Environmental Chemistry

Stoichiometry can be used to determine quantities needed to:

• neutralize acidic waste
• remove pollutants
• treat wastewater
• calculate combustion emissions
• analyze atmospheric reactions

For example:

HCl + NaOH → NaCl + H₂O

The mole ratio between HCl and NaOH is:

1 : 1

Therefore, 1 mol NaOH is stoichiometrically required to neutralize 1 mol HCl.

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Real-World Connection: Combustion and Emissions

Consider:

CH₄ + 2O₂ → CO₂ + 2H₂O

The equation predicts:

1 mol CH₄ → 1 mol CO₂

Therefore, knowing how much methane is burned allows us to calculate the theoretical amount of carbon dioxide produced.

Stoichiometry is therefore important when estimating emissions from chemical processes and fuels.

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Theoretical Amounts vs. Actual Amounts

Stoichiometric calculations predict what should happen according to the balanced equation.

These calculated quantities are theoretical.

Real experiments may produce less product because:

• reactions may not go to completion
• material may be lost during transfer
• competing reactions may occur
• products may be lost during purification
• experimental measurements have uncertainty

Later, these ideas lead to concepts such as:

theoretical yield

and:

percentage yield

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Assumptions in Basic Stoichiometry

Simple stoichiometric calculations often assume:

• the equation is correct and balanced
• reactants are pure
• the reaction proceeds as written
• the required reactants are available
• the reaction proceeds completely
• there are no significant competing reactions
• no product is lost

These assumptions create an idealized calculation.

Actual laboratory results may differ.

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Common Mistakes

Not Balancing the Equation First

Wrong:

H₂ + O₂ → H₂O

Correct:

2H₂ + O₂ → 2H₂O

The mole ratio must come from the balanced equation.

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Using Subscripts as Mole Ratios

For:

2H₂ + O₂ → 2H₂O

the H₂ : O₂ ratio is:

2 : 1

Use coefficients, not subscripts.

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Skipping the Mole Conversion

If a question gives grams, do not usually apply the coefficients directly to the masses.

Use:

grams → moles → mole ratio

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Treating Coefficients as Gram Ratios

For:

2H₂ + O₂ → 2H₂O

the coefficients do not mean:

2 g H₂ + 1 g O₂ → 2 g H₂O

They represent relative numbers of moles.

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Using the Mole Ratio Backwards

Suppose:

N₂ + 3H₂ → 2NH₃

and we are converting H₂ into NH₃.

Use:

2 mol NH₃ / 3 mol H₂

because mol H₂ must cancel.

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Using the Wrong Molar Mass

For O₂:

M = 32.0 g/mol

not 16.0 g/mol.

For CO₂:

M ≈ 44.0 g/mol

not 12.0 g/mol.

Always calculate molar mass for the complete chemical formula.

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Rounding Too Early

Keep several digits during intermediate calculations.

Round appropriately at the end.

Early rounding can make the final answer less accurate.

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Forgetting Units

Always include units.

For example:

0.50 mol

18.0 g

44.0 g/mol

Units help reveal calculation errors.

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Doing Unnecessary Steps

If the question gives moles and asks for moles:

moles → mole ratio → moles

There is no reason to calculate mass first.

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Key Terms

Stoichiometry — The quantitative study of the relationships between reactants and products in chemical reactions.

Stoichiometric calculation — A calculation using a balanced chemical equation to determine quantities of substances.

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

Coefficient — A number before a chemical formula showing the relative amount of that substance in a reaction.

Mole — The amount of substance containing 6.022 × 10²³ representative particles.

Mole ratio — A ratio between amounts of substances obtained from the coefficients of a balanced equation.

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

Reactant — A starting substance in a chemical reaction.

Product — A substance produced by a chemical reaction.

Conversion factor — A ratio used to convert one quantity into another.

Dimensional analysis — A calculation method that uses conversion factors and unit cancellation.

Unit cancellation — The cancellation of identical units appearing in the numerator and denominator of conversion factors.

Conservation of mass — The principle that total mass is conserved during a chemical reaction.

Theoretical amount — The quantity predicted from a balanced chemical equation under ideal conditions.

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

Limiting reactant — The reactant that is consumed first and limits how much product can form.

Excess reactant — A reactant present in more than the amount required by the reaction ratio.

Yield — The amount of product obtained from a chemical reaction.

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Key Takeaways

• Stoichiometry connects quantities of different substances in chemical reactions.
• Every stoichiometric calculation begins with a balanced chemical equation.
• Coefficients provide mole ratios.
• The mole is the central unit connecting different substances.
• A balanced equation directly relates moles, not grams.
• If a problem gives mass, convert mass to moles before using the mole ratio.
• Use n = m/M to convert mass into moles.
• Use m = nM to convert moles into mass.
• Mole-to-mole problems require only the mole ratio.
• Mass-to-mass problems usually require three conversions.
• The standard mass-to-mass pathway is:

mass A → mol A → mol B → mass B

• The conversion between mol A and mol B comes from the balanced equation.
• Unit cancellation can be used to check whether a calculation has been arranged correctly.
• Stoichiometry can calculate reactants from products or products from reactants.
• Multi-step calculations become easier when a conversion pathway is written before calculating.
• Stoichiometric calculations predict theoretical quantities.
• Actual experimental results may differ from theoretical predictions.
• Stoichiometry is used in manufacturing, medicine, environmental science, energy production, and laboratory chemistry.

The most useful roadmap is:

GIVEN QUANTITY → MOLES GIVEN → MOLE RATIO → MOLES WANTED → WANTED QUANTITY

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Check Your Understanding

For questions 1–5, use:

2H₂ + O₂ → 2H₂O

1. How many moles of H₂O can form from 8 mol H₂?

2. How many moles of O₂ are required for 14 mol H₂?

3. How many moles of H₂O can form from 4.5 mol O₂?

4. How many grams of H₂O can form from 2 mol H₂?

5. How many grams of O₂ are required for 5 mol H₂?

For questions 6–10, use:

N₂ + 3H₂ → 2NH₃

6. How many moles of NH₃ can form from 6 mol N₂?

7. How many moles of H₂ are required to produce 10 mol NH₃?

8. How many moles of N₂ are required to produce 16 mol NH₃?

9. How many grams of NH₃ can form from 3 mol N₂? Use M(NH₃) = 17 g/mol.

10. How many grams of H₂ are required to produce 34 g NH₃? Use M(H₂) = 2 g/mol.

For questions 11–15, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

11. How many moles of CO₂ form from 3.5 mol CH₄?

12. How many moles of O₂ are required for 7 mol CH₄?

13. How many grams of CO₂ form from 2 mol CH₄?

14. How many grams of H₂O form from 1.5 mol CH₄?

15. How many moles of CH₄ must burn to produce 88 g CO₂?

For questions 16–20, use:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

16. How many moles of MgO form from 4 mol Mg?

17. How many moles of O₂ are required for 10 mol Mg?

18. How many moles of Mg are present in 48.6 g Mg?

19. How many grams of MgO can theoretically form from 48.6 g Mg?

20. How many grams of Mg are required to produce 80.6 g MgO?

Multi-Step Challenge

Use:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Approximate molar masses:

C₃H₈ = 44 g/mol

O₂ = 32 g/mol

CO₂ = 44 g/mol

H₂O = 18 g/mol

21. How many moles of O₂ are required to burn 2 mol C₃H₈?

22. How many moles of CO₂ form from 5 mol C₃H₈?

23. How many grams of CO₂ form from 44 g C₃H₈?

24. How many grams of H₂O form from 88 g C₃H₈?

25. How many grams of O₂ are required to burn 132 g C₃H₈ completely?

26. How many grams of C₃H₈ must burn to produce 264 g CO₂?

27. Write the complete conversion pathway for converting grams of C₃H₈ into grams of H₂O.

28. Explain why the mole ratio must be taken from a balanced equation.

29. Explain why grams of one substance cannot normally be converted directly into grams of another using the coefficients.

30. A student calculates that 112 g of iron can produce 160 g of iron oxide and claims that mass has been created. Explain why this conclusion is incorrect.