Reacting Quantities

Chemical reactions occur in specific quantitative proportions. A balanced chemical equation tells us not only which substances react, but also the relative amounts required.

For example:

2H₂ + O₂ → 2H₂O

This means:

2 mol H₂ react with 1 mol O₂

Therefore, if we know how much hydrogen is available, we can determine exactly how much oxygen is required.

This is the main idea behind reacting quantities:

balanced equation → mole ratio → required amount

These calculations are extremely useful because chemists rarely want to mix reactants randomly. They want to know how much of each substance is needed for the reaction.

Reactants Must Be Present in the Correct Ratio

Consider:

2H₂ + O₂ → 2H₂O

The required ratio is:

H₂ : O₂ = 2 : 1

So:

• 2 mol H₂ require 1 mol O₂
• 4 mol H₂ require 2 mol O₂
• 6 mol H₂ require 3 mol O₂
• 10 mol H₂ require 5 mol O₂

The quantities change, but the ratio remains:

2 : 1

This is called the stoichiometric ratio.

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Why Balanced Equations Matter

The reacting quantities must come from a balanced chemical equation.

Consider:

Mg + O₂ → MgO

This equation is not balanced.

The balanced equation is:

2Mg + O₂ → 2MgO

Therefore:

2 mol Mg react with 1 mol O₂

If we used the unbalanced equation, we might incorrectly assume:

1 mol Mg reacts with 1 mol O₂

That would give the wrong reacting quantities.

Always:

BALANCE FIRST → CALCULATE SECOND

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Coefficients Give Mole Ratios

Consider:

N₂ + 3H₂ → 2NH₃

The coefficients are:

1 : 3 : 2

Therefore:

1 mol N₂ reacts with 3 mol H₂

and theoretically produces:

2 mol NH₃

For reacting-quantity questions, we often focus on the two reactants:

N₂ : H₂ = 1 : 3

The coefficients provide the conversion factor between them.

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Reacting Quantities in Moles

The simplest reacting-quantity problems give one reactant in moles and ask how many moles of another reactant are required.

The general calculation is:

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

For example:

2H₂ + O₂ → 2H₂O

How many moles of O₂ are required for 8 mol H₂?

8 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 4 mol O₂

Therefore:

8 mol H₂ require 4 mol O₂

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A Simple Calculation Method

For most reacting-quantity problems:

Balance the equation

Make sure the equation is correct.

Identify the known reactant

What quantity has been given?

Identify the required reactant

What quantity must be calculated?

Convert to moles if necessary

If mass is given:

n = m / M

Apply the mole ratio

Use the coefficients from the balanced equation.

Convert to the required unit

If mass is required:

m = nM

The overall pathway is:

KNOWN REACTANT → MOLES → MOLE RATIO → MOLES OF REQUIRED REACTANT → REQUIRED QUANTITY

Worked Example: Hydrogen and Oxygen

Consider:

2H₂ + O₂ → 2H₂O

How many moles of O₂ are required to react completely with 7 mol H₂?

Identify the ratio

H₂ : O₂ = 2 : 1

Calculate

7 mol H₂ × (1 mol O₂ / 2 mol H₂)

= 3.5 mol O₂

Answer

3.5 mol O₂

Notice that reacting quantities do not have to be whole numbers.

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Worked Example: Making Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

How many moles of H₂ are required to react completely with 4 mol N₂?

Ratio:

1 mol N₂ : 3 mol H₂

Calculation:

4 mol N₂ × (3 mol H₂ / 1 mol N₂)

= 12 mol H₂

Answer

12 mol H₂

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Worked Example: Working Backwards

Using:

N₂ + 3H₂ → 2NH₃

How many moles of N₂ are required to react with 15 mol H₂?

Ratio:

1 mol N₂ : 3 mol H₂

Calculation:

15 mol H₂ × (1 mol N₂ / 3 mol H₂)

= 5 mol N₂

Answer

5 mol N₂

Mole ratios can be used in either direction.

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Reacting Quantities Involving Mass

Laboratory chemicals are often measured by mass, not by counting moles directly.

Therefore, many practical questions follow:

mass A → moles A → moles B → mass B

This is one of the most important calculation pathways in chemistry.

Remember:

n = m / M

and:

m = nM

where:

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

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Worked Example: Magnesium and Oxygen

Magnesium reacts with oxygen:

2Mg + O₂ → 2MgO

How many grams of O₂ are required to react completely with 24.3 g Mg?

Molar masses:

Mg = 24.3 g/mol

O₂ = 32.0 g/mol

Convert Mg to moles

n = m / M

n = 24.3 / 24.3

= 1.00 mol Mg

Use the mole ratio

From:

2Mg + O₂ → 2MgO

2 mol Mg : 1 mol O₂

Therefore:

1.00 mol Mg × (1 mol O₂ / 2 mol Mg)

= 0.500 mol O₂

Convert O₂ to mass

m = nM

m = 0.500 × 32.0

= 16.0 g

Answer

16.0 g O₂

The complete pathway was:

24.3 g Mg → 1.00 mol Mg → 0.500 mol O₂ → 16.0 g O₂

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Why Mass Ratios Are Different from Mole Ratios

Consider again:

2Mg + O₂ → 2MgO

The mole ratio is:

2 mol Mg : 1 mol O₂

But this does not mean:

2 g Mg : 1 g O₂

Convert the amounts into mass:

2 mol Mg:

2 × 24.3 = 48.6 g

1 mol O₂:

1 × 32.0 = 32.0 g

Therefore, the reacting mass relationship is:

48.6 g Mg : 32.0 g O₂

Mole ratios and mass ratios are not generally the same.

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Worked Example: Finding the Reacting Mass

Aluminum reacts with chlorine:

2Al + 3Cl₂ → 2AlCl₃

How many grams of chlorine gas are required to react completely with 5.40 g Al?

Use:

Al = 27.0 g/mol

Cl₂ = 71.0 g/mol

Convert aluminum to moles

5.40 ÷ 27.0 = 0.200 mol Al

Apply the mole ratio

Al : Cl₂ = 2 : 3

0.200 mol Al × (3 mol Cl₂ / 2 mol Al)

= 0.300 mol Cl₂

Convert chlorine to mass

0.300 × 71.0 = 21.3 g

Answer

21.3 g Cl₂

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Writing the Calculation as One Line

The same calculation can be written using dimensional analysis:

5.40 g Al × (1 mol Al / 27.0 g Al) × (3 mol Cl₂ / 2 mol Al) × (71.0 g Cl₂ / 1 mol Cl₂)

The units cancel:

g Al → mol Al → mol Cl₂ → g Cl₂

leaving:

21.3 g Cl₂

This method can make complicated stoichiometric calculations easier to organize.

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Worked Example: Iron and Oxygen

Iron reacts with oxygen:

4Fe + 3O₂ → 2Fe₂O₃

How many grams of O₂ are required to react completely with 112 g Fe?

Use:

Fe = 56 g/mol

O₂ = 32 g/mol

Convert Fe to moles

112 ÷ 56 = 2 mol Fe

Use the mole ratio

Fe : O₂ = 4 : 3

2 mol Fe × (3 mol O₂ / 4 mol Fe)

= 1.5 mol O₂

Convert to mass

1.5 × 32 = 48 g

Answer

48 g O₂

Therefore:

112 g Fe reacts with 48 g O₂

If the reaction forms only Fe₂O₃, conservation of mass predicts:

112 g + 48 g = 160 g Fe₂O₃

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Conservation of Mass

Reacting quantities must obey the law of conservation of mass.

For:

4Fe + 3O₂ → 2Fe₂O₃

we found:

112 g Fe + 48 g O₂ → 160 g Fe₂O₃

Total reactant mass:

160 g

Total product mass:

160 g

Mass has been conserved.

Worked Example: Combustion of Methane

Methane burns according to:

CH₄ + 2O₂ → CO₂ + 2H₂O

How many grams of O₂ are required to burn 16 g CH₄ completely?

Use:

CH₄ = 16 g/mol

O₂ = 32 g/mol

Convert methane to moles

16 ÷ 16 = 1 mol CH₄

Use the mole ratio

CH₄ : O₂ = 1 : 2

Therefore:

1 mol CH₄ requires 2 mol O₂

Convert oxygen to mass

2 × 32 = 64 g

Answer

64 g O₂

Therefore:

16 g CH₄ requires 64 g O₂

for complete combustion according to this equation.

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Worked Example: Propane Combustion

Propane burns according to:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

How many grams of oxygen are required to burn 44 g propane?

Use:

C₃H₈ = 44 g/mol

O₂ = 32 g/mol

Convert propane to moles

44 ÷ 44 = 1 mol C₃H₈

Use the ratio

1 mol C₃H₈ : 5 mol O₂

Therefore:

5 mol O₂

Convert to mass

5 × 32 = 160 g

Answer

160 g O₂

So:

44 g propane requires 160 g oxygen

for complete combustion.

Why Combustion Requires So Much Oxygen

Students are sometimes surprised that a relatively small mass of fuel can require a much larger mass of oxygen.

For example:

44 g propane requires 160 g O₂

This happens because combustion combines the fuel with oxygen from the surrounding air.

The mass of the combustion products therefore includes:

mass from the fuel + mass from oxygen

This is why combustion products can have a greater total mass than the original fuel alone.

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Practical Situation: Acid Neutralization

Reacting quantities are important in neutralization.

Consider:

HCl + NaOH → NaCl + H₂O

The mole ratio is:

1 mol HCl : 1 mol NaOH

Therefore:

0.50 mol HCl requires 0.50 mol NaOH

If the molar mass of NaOH is:

40.0 g/mol

then:

m = nM

m = 0.50 × 40.0

= 20.0 g NaOH

Therefore:

0.50 mol HCl requires 20.0 g NaOH

according to the balanced equation.

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Practical Situation: Acid and Carbonate

Consider:

2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

The reacting mole ratio is:

2 mol HCl : 1 mol CaCO₃

Suppose we have:

0.40 mol HCl

How much CaCO₃ is required?

0.40 mol HCl × (1 mol CaCO₃ / 2 mol HCl)

= 0.20 mol CaCO₃

Molar mass of CaCO₃:

100 g/mol

Therefore:

0.20 × 100 = 20 g

Answer

20 g CaCO₃

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Why This Matters in Neutralization

If too little carbonate is used:

some acid remains

If exactly the stoichiometric amount is used:

the reactants are present in the proportion required by the equation.

If more carbonate is added than required:

carbonate remains in excess

This introduces two important ideas:

limiting reactant

and:

excess reactant

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Exact Stoichiometric Quantities

Suppose:

2H₂ + O₂ → 2H₂O

We mix:

4 mol H₂

and:

2 mol O₂

The required ratio is:

2 : 1

Our ratio is:

4 : 2

which simplifies to:

2 : 1

Therefore, the reactants are present in the exact stoichiometric proportion.

If the reaction proceeds completely as written:

• all H₂ can be consumed
• all O₂ can be consumed
• neither is left in excess

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What Happens If the Ratio Is Wrong?

Suppose instead we mix:

4 mol H₂

with:

5 mol O₂

But only:

2 mol O₂

are required for 4 mol H₂.

Therefore:

3 mol O₂ remain

after all the hydrogen has reacted, assuming the reaction proceeds completely.

Hydrogen is the:

limiting reactant

Oxygen is the:

excess reactant

A later topic may examine limiting reactants in more detail, but reacting quantities provide the foundation.

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Another Example of Excess Reactant

Consider:

N₂ + 3H₂ → 2NH₃

Suppose we have:

2 mol N₂

How much H₂ is required?

Ratio:

1 : 3

Therefore:

2 mol N₂ require 6 mol H₂

If we actually supply:

10 mol H₂

then:

6 mol H₂ are required

and:

4 mol H₂ are extra

Hydrogen is present in excess.

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Reacting Quantities and Laboratory Planning

Before performing an experiment, chemists can calculate the required reactant quantities.

Suppose a student wants to react:

0.10 mol Mg

with hydrochloric acid.

Equation:

Mg + 2HCl → MgCl₂ + H₂

The ratio is:

1 mol Mg : 2 mol HCl

Therefore:

0.10 mol Mg requires 0.20 mol HCl

This calculation can be performed before the experiment begins.

Worked Example: Magnesium and Hydrochloric Acid

How many grams of HCl are required to react completely with 4.86 g Mg?

Equation:

Mg + 2HCl → MgCl₂ + H₂

Use:

Mg = 24.3 g/mol

HCl = 36.5 g/mol

Convert Mg to moles

4.86 ÷ 24.3 = 0.200 mol Mg

Apply the mole ratio

Mg : HCl = 1 : 2

0.200 mol Mg × (2 mol HCl / 1 mol Mg)

= 0.400 mol HCl

Convert HCl to mass

0.400 × 36.5 = 14.6 g

Answer

14.6 g HCl

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Practical Chemistry and Safety

Reacting-quantity calculations can also improve laboratory safety.

If chemists calculate quantities before an experiment, they can avoid using unnecessarily large amounts of chemicals.

This can reduce:

• chemical waste
• cost
• exposure to hazardous substances
• quantities requiring disposal
• severity of possible spills

Good stoichiometry therefore supports both:

efficient chemistry

and:

safer chemistry

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Reacting Quantities and Green Chemistry

Using excessive quantities of reactants can create unnecessary waste.

Suppose a reaction requires:

1 mol A : 2 mol B

Using much more B than necessary may:

• waste raw material
• require additional separation
• increase disposal requirements
• increase production costs

Industrial chemists therefore carefully control reacting quantities.

This connects stoichiometry with:

green chemistry and sustainability

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Real-World Application: Fertilizer Production

Ammonia is an important raw material for fertilizer production.

It can be produced using:

N₂ + 3H₂ ⇌ 2NH₃

The stoichiometric relationship is:

1 mol N₂ : 3 mol H₂

Large chemical plants must carefully control the amounts of gases entering industrial processes.

Even though real industrial systems involve additional complications such as equilibrium, recycling, temperature, pressure, and conversion efficiency, the balanced equation provides the basic quantitative relationship.

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Real-World Application: Combustion

Engines, furnaces, boilers, and burners require appropriate amounts of fuel and oxygen.

Too little oxygen can cause:

incomplete combustion

For hydrocarbons, incomplete combustion may produce substances including:

• carbon monoxide
• carbon
• unburned hydrocarbons

Correct reacting quantities therefore have implications for:

• efficiency
• pollution
• fuel consumption
• safety

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Real-World Application: Environmental Treatment

Stoichiometric calculations can help determine how much chemical is required to:

• neutralize acidic waste
• treat alkaline waste
• remove contaminants
• precipitate dissolved substances
• control water chemistry

Using too little treatment chemical may leave contaminants untreated.

Using excessive amounts may:

• waste chemicals
• increase costs
• create additional environmental problems

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Comparing Mole and Mass Relationships

Consider:

2H₂ + O₂ → 2H₂O

Mole relationship

2 mol H₂ : 1 mol O₂

Using molar masses:

H₂ = 2 g/mol

O₂ = 32 g/mol

Mass relationship

2 mol H₂:

2 × 2 = 4 g

1 mol O₂:

1 × 32 = 32 g

Therefore:

4 g H₂ reacts with 32 g O₂

Notice:

mole ratio = 2 : 1

but:

mass ratio = 4 : 32 = 1 : 8

These ratios are very different.

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Scaling Reacting Quantities

Once we know the correct reacting quantities, we can scale them.

For:

4 g H₂ : 32 g O₂

divide both by 4:

1 g H₂ : 8 g O₂

Multiply by 10:

10 g H₂ : 80 g O₂

Multiply by 25:

25 g H₂ : 200 g O₂

The mass ratio remains constant because it comes from the stoichiometric mole relationship and the substances' molar masses.

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Worked Example: Scaling by Mass

Suppose:

2H₂ + O₂ → 2H₂O

We know:

4 g H₂ requires 32 g O₂

How much oxygen is required for:

12 g H₂?

12 g is three times 4 g.

Therefore:

32 × 3 = 96 g O₂

Answer

96 g O₂

This proportional method works when the reacting mass relationship is already known.

For unfamiliar reactions, the mole method is generally safer.

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A Reliable Problem-Solving Checklist

Before calculating, ask:

Is the equation balanced?

Then identify:

What substance do I know?

What substance do I need?

What unit was I given?

What unit is required?

Then write the pathway:

given → mol given → mol wanted → wanted unit

Finally ask:

Does my answer make chemical sense?

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Worked Example: Full Multi-Step Problem

Calcium reacts with water:

Ca + 2H₂O → Ca(OH)₂ + H₂

How many grams of water are required to react completely with 20.0 g Ca?

Use:

Ca = 40.0 g/mol

H₂O = 18.0 g/mol

Convert calcium to moles

20.0 ÷ 40.0 = 0.500 mol Ca

Apply the mole ratio

Ca : H₂O = 1 : 2

0.500 mol Ca × (2 mol H₂O / 1 mol Ca)

= 1.00 mol H₂O

Convert water to mass

1.00 × 18.0 = 18.0 g

Answer

18.0 g H₂O

The pathway was:

20.0 g Ca → 0.500 mol Ca → 1.00 mol H₂O → 18.0 g H₂O

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Worked Example: A More Challenging Reaction

Consider:

2Al + 3CuCl₂ → 2AlCl₃ + 3Cu

How many grams of CuCl₂ are required to react completely with 5.40 g Al?

Use:

Al = 27.0 g/mol

CuCl₂ = 134.5 g/mol

Convert Al to moles

5.40 ÷ 27.0 = 0.200 mol Al

Apply the ratio

Al : CuCl₂ = 2 : 3

0.200 mol Al × (3 mol CuCl₂ / 2 mol Al)

= 0.300 mol CuCl₂

Convert to mass

0.300 × 134.5 = 40.35 g

Answer

Approximately:

40.4 g CuCl₂

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Checking the Answer

After completing a calculation, check:

Equation

Was it balanced?

Mole Ratio

Did you use coefficients rather than subscripts?

Direction

Did the given substance cancel?

Molar Mass

Did you calculate the complete formula correctly?

Units

Does your final answer have the requested unit?

Magnitude

Does the answer seem reasonable?

These checks catch many common errors.

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Common Mistakes

Using an Unbalanced Equation

Mole ratios only work correctly with balanced equations.

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Treating Mole Ratios as Mass Ratios

For:

2H₂ + O₂ → 2H₂O

2 : 1 is a mole ratio, not a gram ratio.

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Forgetting to Convert Mass to Moles

If mass is given, usually begin:

mass → moles

before using the coefficients.

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Using Subscripts Instead of Coefficients

For:

2Mg + O₂ → 2MgO

Mg : O₂ = 2 : 1

Do not use the subscripts in the formulas to create the mole ratio.

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Reversing the Mole Ratio

If converting Mg into O₂:

2Mg + O₂ → 2MgO

use:

1 mol O₂ / 2 mol Mg

so that mol Mg cancels.

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Using Atomic Mass Instead of Molecular Molar Mass

Oxygen gas is:

O₂

Therefore:

M(O₂) = 32.0 g/mol

not 16.0 g/mol.

Similarly:

Cl₂ ≈ 71.0 g/mol

not 35.5 g/mol.

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Assuming Equal Masses React

Equal numbers of moles do not necessarily have equal masses.

Different substances have different molar masses.

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Assuming More Reactant Is Always Better

Excess reactant may:

• waste material
• increase cost
• require separation
• increase waste

The correct quantity depends on the reaction and purpose.

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Forgetting Conservation of Mass

If a product has more mass than one reactant, that does not mean mass was created.

Other reactants contributed mass.

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Rounding Too Early

Keep extra digits during intermediate calculations and round at the end.

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Key Terms

Reacting quantity — The amount of a substance required or involved in a chemical reaction.

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

Stoichiometric ratio — The mole relationship between substances specified by a balanced equation.

Stoichiometric amount — The amount of a substance required according to the balanced chemical equation.

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

Coefficient — A number before a chemical formula indicating its relative amount in the reaction.

Mole ratio — A ratio between amounts in moles obtained from coefficients in 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.

Product — A substance formed during a chemical reaction.

Conversion factor — A ratio used to convert between quantities.

Dimensional analysis — A method of calculation using conversion factors and unit cancellation.

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

Limiting reactant — The reactant that is consumed first and therefore limits product formation.

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

Complete combustion — Combustion in sufficient oxygen that, for a hydrocarbon, ideally produces carbon dioxide and water.

Incomplete combustion — Combustion occurring with insufficient oxygen, potentially producing carbon monoxide, carbon, and other products.

Neutralization — A reaction in which an acid and base react, typically producing a salt and water.

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Key Takeaways

• Chemical reactions require reactants in specific proportions.
• These proportions come from balanced chemical equations.
• Coefficients provide the mole ratios between reactants.
• Reacting quantities are fundamentally based on moles.
• A balanced equation must be used before any stoichiometric calculation.
• If one reacting quantity is known, the required quantity of another reactant can be calculated.
• For mole-to-mole problems:

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

• For mass-to-mass reacting-quantity problems:

mass A → mol A → mol B → mass B

• Convert mass to moles using:

n = m/M

• Convert moles to mass using:

m = nM

• Mole ratios are not usually the same as mass ratios.
• Different substances have different molar masses.
• Unit cancellation helps verify that calculations are arranged correctly.
• Reacting quantities can be scaled while maintaining the same stoichiometric proportions.
• If reactants are supplied in exactly the required ratio, neither should remain in excess after complete reaction as written.
• If one reactant is supplied in excess, another reactant limits how far the reaction can proceed.
• Conservation of mass applies to all reacting quantities.
• Practical stoichiometry helps reduce waste, control costs, and improve laboratory safety.
• Reacting-quantity calculations are used in combustion, neutralization, manufacturing, environmental treatment, and many other chemical processes.

The central strategy is:

BALANCE → CONVERT TO MOLES → USE THE REACTANT MOLE RATIO → CONVERT TO THE REQUIRED QUANTITY

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Check Your Understanding

For questions 1–5, use:

2H₂ + O₂ → 2H₂O

1. How many moles of O₂ are required for 6 mol H₂?

2. How many moles of H₂ are required for 4 mol O₂?

3. How many grams of O₂ are required for 4 mol H₂?

4. How many grams of H₂ are required to react with 64 g O₂? Use M(H₂) = 2.0 g/mol.

5. Explain why the mole ratio 2 : 1 is not the same as the reacting mass ratio.

For questions 6–10, use:

N₂ + 3H₂ → 2NH₃

6. How many moles of H₂ are required for 5 mol N₂?

7. How many moles of N₂ are required for 21 mol H₂?

8. How many grams of H₂ are required for 2 mol N₂?

9. How many moles of H₂ are required for 28 g N₂? Use M(N₂) = 28 g/mol.

10. How many grams of N₂ are required to react with 12 g H₂?

For questions 11–15, use:

2Mg + O₂ → 2MgO

Use:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

11. How many moles of O₂ are required for 6 mol Mg?

12. How many grams of O₂ are required for 48.6 g Mg?

13. How many grams of Mg are required to react with 16.0 g O₂?

14. A student mixes 4 mol Mg with 2 mol O₂. Are the reactants in the correct stoichiometric proportion? Explain.

15. A student mixes 4 mol Mg with 5 mol O₂. Which substance is present in excess?

For questions 16–20, use:

CH₄ + 2O₂ → CO₂ + 2H₂O

Use:

M(CH₄) = 16 g/mol

M(O₂) = 32 g/mol

16. How many moles of O₂ are required for 3 mol CH₄?

17. How many grams of O₂ are required to burn 16 g CH₄?

18. How many grams of CH₄ can react completely with 128 g O₂?

19. Explain why 16 g CH₄ requires a much greater mass of oxygen.

20. Why can insufficient oxygen change the products formed during combustion?

Multi-Step Challenge

Use:

2Al + 3Cl₂ → 2AlCl₃

Molar masses:

Al = 27.0 g/mol

Cl₂ = 71.0 g/mol

21. How many moles of Cl₂ are required for 4 mol Al?

22. How many grams of Cl₂ are required for 2 mol Al?

23. How many moles of Al are required for 6 mol Cl₂?

24. How many grams of Cl₂ are required to react with 10.8 g Al?

25. How many grams of Al are required to react with 35.5 g Cl₂?

26. Write the complete conversion pathway for calculating grams of Cl₂ required from grams of Al.

27. Explain why coefficients rather than subscripts determine reacting quantities.

28. Explain why chemists calculate reacting quantities before performing laboratory experiments.

29. Explain how accurate reacting-quantity calculations can reduce chemical waste.

30. A factory uses much more of one reactant than the balanced equation requires. Explain two possible disadvantages of doing this.

Extended Challenge

Calcium carbonate reacts with hydrochloric acid:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Use:

M(CaCO₃) = 100 g/mol

M(HCl) = 36.5 g/mol

31. How many moles of HCl are required for 1 mol CaCO₃?

32. How many moles of HCl are required for 2.5 mol CaCO₃?

33. How many grams of HCl are required for 100 g CaCO₃?

34. How many grams of CaCO₃ are required to react with 73 g HCl?

35. A student has 50 g CaCO₃. Calculate the mass of HCl required for complete reaction.

36. Explain what would happen if the student used less HCl than the calculated amount.

37. Explain what would happen if considerably more HCl were added than required.

38. Identify which reactant would be in excess in Question 37.

39. Explain how this reaction demonstrates the connection between mole ratios and practical reacting quantities.

40. Explain why the pathway

mass CaCO₃ → mol CaCO₃ → mol HCl → mass HCl

is more reliable than simply comparing the masses of CaCO₃ and HCl directly.