Predicting Product Amounts

One of the most useful applications of stoichiometry is predicting how much product can form from a known amount of reactant.

A balanced chemical equation provides the relationship between reactants and products.

For example:

2H₂ + O₂ → 2H₂O

This tells us:

2 mol H₂ → 2 mol H₂O

Therefore, if enough oxygen is available:

5 mol H₂ → 5 mol H₂O

The general idea is:

known reactant → balanced equation → predicted product

The calculated product amount represents the amount expected from the chemical equation under the assumptions given in the problem.

Balanced Equations Predict Product Quantities

Consider:

N₂ + 3H₂ → 2NH₃

The coefficients tell us:

1 mol N₂ + 3 mol H₂ → 2 mol NH₃

Therefore:

• 1 mol N₂ can produce 2 mol NH₃
• 2 mol N₂ can produce 4 mol NH₃
• 5 mol N₂ can produce 10 mol NH₃
• 10 mol N₂ can produce 20 mol NH₃

These predictions assume that enough hydrogen is available.

The equation provides the stoichiometric relationship between the reactant and product.

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The Main Calculation Pathway

When predicting product amounts, use:

KNOWN REACTANT → MOLES OF REACTANT → MOLE RATIO → MOLES OF PRODUCT → REQUIRED PRODUCT UNIT

If both quantities are measured in moles:

mol reactant → mol product

If the reactant is given in grams and the product is required in grams:

g reactant → mol reactant → mol product → g product

This second pathway is one of the most important calculations in stoichiometry.

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Predicting Product in Moles

Consider:

2Mg + O₂ → 2MgO

How many moles of MgO can form from 7 mol Mg, assuming sufficient oxygen?

Identify the mole ratio

Mg : MgO = 2 : 2

This simplifies to:

1 : 1

Calculate

7 mol Mg × (2 mol MgO / 2 mol Mg)

= 7 mol MgO

Answer

7 mol MgO

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A Useful Formula

For mole-to-mole product calculations:

moles product = moles reactant × (coefficient product / coefficient reactant)

For example:

4Fe + 3O₂ → 2Fe₂O₃

If we begin with 8 mol Fe:

moles Fe₂O₃ = 8 × (2/4)

= 4 mol Fe₂O₃

This formula works only when the equation is balanced and the selected reactant is available to react as assumed.

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Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

How many moles of ammonia can theoretically form from 6 mol N₂, assuming sufficient hydrogen?

Mole ratio

N₂ : NH₃ = 1 : 2

Calculate

6 mol N₂ × (2 mol NH₃ / 1 mol N₂)

= 12 mol NH₃

Answer

12 mol NH₃

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Worked Example: Starting with Hydrogen

Using:

N₂ + 3H₂ → 2NH₃

How many moles of NH₃ can form from 9 mol H₂, assuming sufficient nitrogen?

Ratio:

3 mol H₂ : 2 mol NH₃

Calculation:

9 mol H₂ × (2 mol NH₃ / 3 mol H₂)

= 6 mol NH₃

Answer

6 mol NH₃

The product prediction depends on which reactant quantity is given.

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Predicting Product Mass

Laboratory quantities are often measured in grams.

To calculate product mass:

mass reactant → moles reactant → moles product → mass product

Use:

n = m/M

to convert mass to moles.

Then use the mole ratio.

Finally use:

m = nM

to convert product moles to product mass.

Worked Example: Magnesium Oxide

Magnesium burns in oxygen:

2Mg + O₂ → 2MgO

What mass of MgO can form from 24.3 g Mg, assuming sufficient oxygen?

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

Convert magnesium to moles

n = m/M

n = 24.3 / 24.3

= 1.00 mol Mg

Convert Mg to MgO

Mg : MgO = 1 : 1

Therefore:

1.00 mol MgO

Convert MgO to mass

m = nM

m = 1.00 × 40.3

= 40.3 g

Answer

40.3 g MgO

The pathway was:

24.3 g Mg → 1.00 mol Mg → 1.00 mol MgO → 40.3 g MgO

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Why the Product Has More Mass

In the previous example:

24.3 g Mg → 40.3 g MgO

It may appear that mass has been created.

It has not.

Magnesium combines with oxygen:

2Mg + O₂ → 2MgO

The additional mass comes from oxygen.

For 1 mol Mg:

24.3 g Mg + 16.0 g O → 40.3 g MgO

The total mass is conserved.

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Worked Example: Predicting Water

Hydrogen burns according to:

2H₂ + O₂ → 2H₂O

What mass of water can form from 6.0 g H₂, assuming sufficient oxygen?

Use:

M(H₂) = 2.0 g/mol

M(H₂O) = 18.0 g/mol

Convert hydrogen to moles

6.0 ÷ 2.0 = 3.0 mol H₂

Apply the mole ratio

H₂ : H₂O = 2 : 2 = 1 : 1

Therefore:

3.0 mol H₂O

Convert water to mass

3.0 × 18.0 = 54 g

Answer

54 g H₂O

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Worked Example: Iron Oxide

Iron reacts with oxygen:

4Fe + 3O₂ → 2Fe₂O₃

What mass of Fe₂O₃ can form from 56 g Fe, assuming sufficient oxygen?

Use:

M(Fe) = 56 g/mol

M(Fe₂O₃) = 160 g/mol

Convert iron to moles

56 ÷ 56 = 1 mol Fe

Use the mole ratio

Fe : Fe₂O₃ = 4 : 2

1 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 0.5 mol Fe₂O₃

Convert to mass

0.5 × 160 = 80 g

Answer

80 g Fe₂O₃

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Dimensional Analysis

The iron oxide calculation can also be written as one continuous calculation:

56 g Fe × (1 mol Fe / 56 g Fe) × (2 mol Fe₂O₃ / 4 mol Fe) × (160 g Fe₂O₃ / 1 mol Fe₂O₃)

Units cancel:

g Fe → mol Fe → mol Fe₂O₃ → g Fe₂O₃

leaving:

80 g Fe₂O₃

This method is useful for checking that each conversion has been arranged correctly.

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Predicting Products from Combustion

Methane undergoes complete combustion:

CH₄ + 2O₂ → CO₂ + 2H₂O

From the equation:

1 mol CH₄ → 1 mol CO₂

and:

1 mol CH₄ → 2 mol H₂O

Therefore, burning one substance can produce different mole quantities of different products.

Worked Example: Carbon Dioxide from Methane

How many grams of CO₂ can form when 32 g CH₄ burns completely?

Use:

M(CH₄) = 16 g/mol

M(CO₂) = 44 g/mol

Convert methane to moles

32 ÷ 16 = 2 mol CH₄

Apply the mole ratio

CH₄ : CO₂ = 1 : 1

Therefore:

2 mol CO₂

Convert to mass

2 × 44 = 88 g

Answer

88 g CO₂

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Worked Example: Water from Methane

Using the same reaction:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many grams of water can form from 32 g CH₄?

We already know:

32 g CH₄ = 2 mol CH₄

Ratio:

1 mol CH₄ : 2 mol H₂O

Therefore:

2 mol CH₄ × (2 mol H₂O / 1 mol CH₄)

= 4 mol H₂O

Convert to mass:

4 × 18 = 72 g

Answer

72 g H₂O

So complete combustion of 32 g methane theoretically produces:

88 g CO₂

and:

72 g H₂O

assuming sufficient oxygen.

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Checking Conservation of Mass

For:

CH₄ + 2O₂ → CO₂ + 2H₂O

Suppose:

32 g CH₄

reacts.

This is:

2 mol CH₄

It requires:

4 mol O₂

Mass of oxygen:

4 × 32 = 128 g

Total reactant mass:

32 + 128 = 160 g

Predicted products:

88 g CO₂ + 72 g H₂O = 160 g

Therefore:

mass of reactants = mass of products

The prediction agrees with conservation of mass.

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Predicting Multiple Products

Some reactions produce more than one product.

Consider:

CaCO₃ → CaO + CO₂

One mole of calcium carbonate produces:

1 mol CaO

and:

1 mol CO₂

Suppose 2 mol CaCO₃ decompose completely.

We predict:

2 mol CaO

and:

2 mol CO₂

Each product can then be converted into mass if required.

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Worked Example: Thermal Decomposition

How much CO₂ can form from 250 g CaCO₃?

Equation:

CaCO₃ → CaO + CO₂

Use:

M(CaCO₃) = 100 g/mol

M(CO₂) = 44 g/mol

Convert CaCO₃ to moles

250 ÷ 100 = 2.5 mol CaCO₃

Apply the mole ratio

CaCO₃ : CO₂ = 1 : 1

Therefore:

2.5 mol CO₂

Convert to mass

2.5 × 44 = 110 g

Answer

110 g CO₂

Predicting the Other Product

Using:

CaCO₃ → CaO + CO₂

What mass of CaO forms from the same 250 g CaCO₃?

We already know:

250 g CaCO₃ = 2.5 mol CaCO₃

The ratio CaCO₃ : CaO is:

1 : 1

Therefore:

2.5 mol CaO

Use:

M(CaO) = 56 g/mol

Mass:

2.5 × 56 = 140 g

Answer

140 g CaO

Now check:

140 g CaO + 110 g CO₂ = 250 g

The predicted products equal the original reactant mass.

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Predicting Product from an Acid Reaction

Magnesium reacts with hydrochloric acid:

Mg + 2HCl → MgCl₂ + H₂

How many moles of hydrogen gas can form from 0.30 mol Mg, assuming sufficient HCl?

Ratio:

1 mol Mg : 1 mol H₂

Therefore:

0.30 mol Mg → 0.30 mol H₂

Answer

0.30 mol H₂

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Predicting Product Mass from an Acid Reaction

Suppose 12.15 g Mg reacts with sufficient hydrochloric acid.

Equation:

Mg + 2HCl → MgCl₂ + H₂

Use:

M(Mg) = 24.3 g/mol

M(MgCl₂) ≈ 95.3 g/mol

Convert magnesium to moles

12.15 ÷ 24.3 = 0.500 mol Mg

Apply the mole ratio

Mg : MgCl₂ = 1 : 1

Therefore:

0.500 mol MgCl₂

Convert to mass

0.500 × 95.3 = 47.65 g

Answer

Approximately:

47.7 g MgCl₂

The additional mass comes from chlorine supplied by hydrochloric acid.

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Predicting Product from a Precipitation Reaction

Consider:

AgNO₃ + NaCl → AgCl + NaNO₃

Silver chloride, AgCl, forms as a solid precipitate.

The mole ratio is:

1 mol AgNO₃ : 1 mol AgCl

If 0.25 mol AgNO₃ reacts with sufficient NaCl:

0.25 mol AgCl

is predicted to form.

If:

M(AgCl) ≈ 143.5 g/mol

then:

m = 0.25 × 143.5

≈ 35.9 g AgCl

Predicted amount

35.9 g AgCl

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Product Predictions Are Theoretical

Stoichiometric calculations tell us how much product should form according to the balanced equation and assumptions of the problem.

This is sometimes called the theoretical yield.

For example, a calculation may predict:

25.0 g product

But an experiment might actually produce:

21.8 g product

The calculation is not necessarily wrong.

Real reactions are not always perfectly efficient.

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Why Actual Product Amounts May Be Lower

The actual amount of product may be lower because:

• the reaction does not go to completion
• product is lost during transfer
• product remains in laboratory equipment
• competing reactions occur
• reactants contain impurities
• some product is lost during filtration
• some product is lost during heating or purification
• measurement uncertainty affects results

This means:

predicted amount ≠ always actual amount

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Theoretical Yield

The theoretical yield is the maximum amount of product predicted by stoichiometry from the available reactant under the stated assumptions.

For example:

Calculation predicts:

12.5 g Cu

Therefore:

theoretical yield = 12.5 g Cu

If only 10.8 g is collected experimentally:

actual yield = 10.8 g Cu

These values can later be used to calculate percentage yield.

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Is a Prediction Reasonable?

A calculation should never end with simply writing a number.

Ask whether the answer is reasonable.

Useful checks include:

• Is the equation balanced?
• Did I use the correct mole ratio?
• Did my units cancel correctly?
• Did I use the correct molar masses?
• Does the size of the answer make sense?
• Does the answer agree with conservation of mass?
• Did I accidentally use coefficients as mass ratios?
• Did I round too early?

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Reasonableness Check: Mole Ratio

Consider:

2Al + 3Cl₂ → 2AlCl₃

If we start with:

4 mol Al

we should produce:

4 mol AlCl₃

because Al : AlCl₃ is:

2 : 2 = 1 : 1

If someone calculates:

12 mol AlCl₃

we should immediately question the result.

The balanced equation provides a quick estimate before detailed calculations begin.

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Reasonableness Check: Conservation of Mass

Suppose a reaction has:

20 g of reactant A

and:

30 g of reactant B

and both react completely to form one product.

The product cannot have a mass of:

80 g

because only:

50 g

of reactants were present.

If no matter enters or leaves the system:

maximum total product mass = 50 g

Conservation of mass provides an important check.

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Reasonableness Check: Order of Magnitude

Suppose:

1 mol reactant → 1 mol product

and both substances have similar molar masses.

If you begin with approximately:

10 g reactant

but calculate:

10,000 g product

something is probably wrong.

Possible errors include:

• incorrect molar mass
• incorrect units
• inverted conversion factor
• calculator entry error

Estimating before calculating helps identify these mistakes.

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The Reactant Used for the Prediction Matters

Consider:

2H₂ + O₂ → 2H₂O

Suppose we have:

10 mol H₂

and:

2 mol O₂

Using hydrogen alone would predict:

10 mol H₂O

But using oxygen predicts:

4 mol H₂O

Both cannot be produced.

Why?

There is not enough oxygen to react with all the hydrogen.

The reactant that runs out first controls the maximum product amount.

This reactant is called the limiting reactant.

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A Preview of Limiting Reactants

For:

2H₂ + O₂ → 2H₂O

Given:

10 mol H₂

and:

2 mol O₂

The oxygen can react with:

4 mol H₂

and produce:

4 mol H₂O

Therefore:

O₂ is the limiting reactant

and:

H₂ is in excess

The maximum product is:

4 mol H₂O

This is why product predictions require careful attention when quantities of both reactants are provided.

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Worked Example: Product Prediction with Two Reactants

Consider:

N₂ + 3H₂ → 2NH₃

Suppose:

2 mol N₂

and:

9 mol H₂

are available.

For 2 mol N₂, the required H₂ is:

2 × 3 = 6 mol H₂

But 9 mol H₂ are available.

Therefore, hydrogen is available in excess.

The 2 mol N₂ determine the product amount.

Ratio:

1 mol N₂ : 2 mol NH₃

Therefore:

2 mol N₂ → 4 mol NH₃

Maximum predicted product

4 mol NH₃

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Industrial Product Predictions

Chemical manufacturers must predict how much product can be made from available raw materials.

Stoichiometric predictions help determine:

• required reactant quantities
• expected production
• raw-material costs
• equipment requirements
• storage requirements
• waste production
• process efficiency

Large industrial processes may involve thousands or millions of kilograms of material, making accurate calculations extremely important.

Application: Ammonia Production

Ammonia can be produced using:

N₂ + 3H₂ ⇌ 2NH₃

Suppose an idealized calculation begins with:

100 mol N₂

and sufficient hydrogen.

The ratio is:

1 mol N₂ : 2 mol NH₃

Therefore:

100 mol N₂ → 200 mol NH₃

The stoichiometric prediction is:

200 mol NH₃

Real industrial production is more complicated because the reaction is reversible and does not simply convert every molecule in a single pass, but the balanced equation still provides the fundamental quantitative relationship.

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Application: Environmental Chemistry

Product predictions can help estimate quantities of substances released into the environment.

For example:

CH₄ + 2O₂ → CO₂ + 2H₂O

The equation predicts:

1 mol CH₄ → 1 mol CO₂

Therefore, the amount of methane burned can be used to estimate the theoretical amount of carbon dioxide produced during complete combustion.

Similar calculations can be applied to:

• fuel combustion
• industrial emissions
• waste treatment
• neutralization
• water treatment

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Application: Laboratory Planning

Suppose a laboratory investigation requires approximately:

5.0 g of a product

Before performing the experiment, a chemist can work backward using stoichiometry to determine how much reactant should theoretically be required.

This helps:

• reduce waste
• control costs
• choose appropriate equipment
• improve safety
• plan experiments efficiently

Stoichiometry therefore allows us to predict both:

reactant → product

and:

desired product → required reactant

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Common Mistakes

Using an Unbalanced Equation

Always balance before calculating.

Wrong:

H₂ + O₂ → H₂O

Correct:

2H₂ + O₂ → 2H₂O

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Using Subscripts Instead of Coefficients

Mole ratios come from coefficients.

For:

2Mg + O₂ → 2MgO

Mg : MgO is:

2 : 2

not a ratio taken from the subscripts.

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Treating Coefficients as Mass Ratios

For:

2H₂ + O₂ → 2H₂O

2 mol H₂ produce 2 mol H₂O.

But:

4 g H₂ → 36 g H₂O

not:

2 g H₂ → 2 g H₂O

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Skipping the Mole Step

For mass-to-mass calculations, use:

mass reactant → mol reactant → mol product → mass product

Do not normally jump directly from one mass to another using coefficients.

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Using the Wrong Molar Mass

For example:

CO₂ = 12 + (2 × 16) = 44 g/mol

Make sure every atom in the formula is included.

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Reversing the Mole Ratio

If converting Fe into Fe₂O₃:

4Fe + 3O₂ → 2Fe₂O₃

use:

2 mol Fe₂O₃ / 4 mol Fe

so mol Fe cancels.

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Assuming Predicted Yield Equals Actual Yield

Stoichiometry gives a theoretical prediction.

Actual laboratory yield may be lower.

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Ignoring the Other Reactant

If quantities of two reactants are given, do not automatically calculate product from whichever appears first.

One may be the limiting reactant.

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Rejecting an Answer Because Product Mass Is Larger

A product can have more mass than one individual reactant because mass from another reactant has been added.

Always consider the total mass of all reactants.

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Rounding Too Early

Keep extra digits during intermediate calculations.

Round the final answer appropriately.

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Key Terms

Product — A substance formed during a chemical reaction.

Predicted product amount — The quantity of product calculated from stoichiometry.

Stoichiometry — The quantitative study of relationships between reactants and products.

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

Coefficient — A number placed before a chemical formula indicating relative amounts in a reaction.

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

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

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

Reactant — A starting substance in a chemical reaction.

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

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

Conservation of mass — The principle that total mass remains constant during a chemical reaction.

Theoretical yield — The maximum quantity of product predicted by stoichiometric calculations under the stated assumptions.

Actual yield — The quantity of product actually obtained experimentally.

Limiting reactant — The reactant consumed first, which determines the maximum amount of product that can form.

Excess reactant — A reactant present in more than the amount required to react with the limiting reactant.

Complete reaction — A reaction in which the relevant reactant is assumed to react as fully as the problem specifies.

Reasonableness check — An evaluation of whether a calculated result is consistent with chemical principles, units, ratios, and expected magnitude.

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Key Takeaways

• Balanced equations can be used to predict quantities of products.
• Product calculations are based on mole ratios.
• Mole ratios come from coefficients in balanced equations.
• For mole-to-mole calculations:

mol reactant → mol product

• For mass-to-mass calculations:

g reactant → mol reactant → mol product → g product

• Convert mass to moles using:

n = m/M

• Convert moles to mass using:

m = nM

• Product mass can be greater than the mass of one reactant because other reactants contribute mass.
• Total mass must still obey conservation of mass.
• Predicted product quantities represent theoretical results under the assumptions of the calculation.
• Actual experimental amounts may be lower than predicted amounts.
• Product predictions should always be checked for reasonableness.
• Unit cancellation is useful for checking calculations.
• Conservation of mass provides another useful check.
• When quantities of multiple reactants are given, the limiting reactant determines the maximum product amount.
• Stoichiometric predictions are important in laboratories, manufacturing, environmental science, energy production, and many other applications.

The main pathway to remember is:

KNOWN REACTANT → MOLES REACTANT → MOLE RATIO → MOLES PRODUCT → PRODUCT 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 6 mol H₂?

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

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

4. How many grams of H₂O can form from 32 g O₂?

5. Explain why the mass of water produced can be greater than the mass of hydrogen used.

For questions 6–10, use:

N₂ + 3H₂ → 2NH₃

6. How many moles of NH₃ can form from 4 mol N₂?

7. How many moles of NH₃ can form from 12 mol H₂?

8. How many grams of NH₃ can form from 2 mol N₂? Use M(NH₃) = 17 g/mol.

9. How many grams of NH₃ can theoretically form from 28 g N₂?

10. Explain why these calculations represent theoretical predictions.

For questions 11–15, use:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(MgO) = 40.3 g/mol

11. How many moles of MgO can form from 5 mol Mg?

12. How many grams of MgO can form from 2 mol Mg?

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

14. A student predicts 40.3 g MgO from 24.3 g Mg. Explain why this does not violate conservation of mass.

15. Identify the additional reactant contributing mass to MgO.

For questions 16–20, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

16. How many moles of CO₂ form from 5 mol CH₄?

17. How many moles of H₂O form from 5 mol CH₄?

18. How many grams of CO₂ form from 32 g CH₄?

19. How many grams of H₂O form from 16 g CH₄?

20. Why are the masses of CO₂ and H₂O produced not equal even though they come from the same reaction?

Multi-Step Challenge

Use:

4Fe + 3O₂ → 2Fe₂O₃

Use:

M(Fe) = 56 g/mol

M(Fe₂O₃) = 160 g/mol

21. How many moles of Fe₂O₃ can form from 8 mol Fe?

22. How many grams of Fe₂O₃ can form from 4 mol Fe?

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

24. Write the complete conversion pathway for Question 23.

25. Explain why product mass is greater than the mass of iron used.

Use:

CaCO₃ → CaO + CO₂

with:

M(CaCO₃) = 100 g/mol

M(CaO) = 56 g/mol

M(CO₂) = 44 g/mol

26. How many moles of CO₂ form from 3 mol CaCO₃?

27. How many grams of CaO form from 200 g CaCO₃?

28. How many grams of CO₂ form from 200 g CaCO₃?

29. Add your answers to Questions 27 and 28. How does the result compare with the original mass of CaCO₃?

30. Explain how Question 29 demonstrates conservation of mass.

Reasonableness Challenge

31. A student calculates that 1 mol Mg produces 10 mol MgO from:

2Mg + O₂ → 2MgO

Explain why the answer cannot be correct.

32. A calculation predicts 5000 g of product from a total of 50 g of reactants in a closed system. Explain why the prediction is unreasonable.

33. A student predicts 36 g H₂O from 4 g H₂ reacting with sufficient O₂. Explain why the product can have a greater mass than the hydrogen.

34. A reaction is predicted to produce 25.0 g of product, but only 21.2 g is collected. Give three possible reasons for the difference.

35. Explain the difference between a predicted theoretical amount and an actual experimental amount.

36. Why should a chemist estimate the approximate size of an answer before completing a stoichiometric calculation?

37. Explain how units can help identify an incorrectly arranged calculation.

38. Explain why the balanced equation is essential for predicting product amounts.

39. Describe how product predictions could help a chemical manufacturer plan production.

40. Explain why identifying the limiting reactant becomes important when quantities of two or more reactants are provided.