Solution Stoichiometry

Solution stoichiometry combines concentration calculations with the mole ratios in balanced chemical equations.

When a reactant is dissolved in a solution, we may not be given its mass directly. Instead, we are often given:

• the concentration of the solution
• the volume of the solution

From these, we can calculate the number of moles:

n = cV

where:

• n = amount of substance in moles
• c = molar concentration in mol/L
• V = volume in litres

Once the number of moles is known, the balanced chemical equation tells us how many moles of another reactant or product are involved.

The basic pathway is:

concentration + volume → moles → mole ratio → required quantity

Why Solution Stoichiometry Is Useful

Many chemical reactions occur in solution.

Examples include:

• acid-base reactions
• precipitation reactions
• reactions involving dissolved ionic compounds
• analytical chemistry
• environmental testing
• industrial chemical processes

Instead of weighing every reactant, chemists can measure a known volume of a solution with a known concentration.

Because:

n = cV

the amount of reactant can be determined accurately.

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The Core Calculation

Suppose we have:

250 mL of 0.40 mol/L NaOH

How many moles of NaOH are present?

First convert the volume:

250 mL = 0.250 L

Then:

n = cV

n = 0.40 × 0.250

n = 0.100 mol

Therefore:

250 mL of 0.40 mol/L NaOH contains 0.100 mol NaOH.

This is usually the first step in solution stoichiometry.

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The Importance of the Balanced Equation

Consider:

HCl + NaOH → NaCl + H₂O

The coefficients show:

1 mol HCl : 1 mol NaOH : 1 mol NaCl : 1 mol H₂O

Therefore:

0.10 mol HCl

requires:

0.10 mol NaOH

and produces:

0.10 mol NaCl

The balanced equation provides the bridge between different substances.

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The Solution Stoichiometry Pathway

Most problems can be solved using the following sequence:

Step 1: Write or identify the balanced equation

For example:

HCl + NaOH → NaCl + H₂O

Step 2: Convert volume to litres

For example:

200 mL = 0.200 L

Step 3: Calculate moles

Use:

n = cV

Step 4: Apply the mole ratio

Use the coefficients from the balanced equation.

Step 5: Calculate the requested quantity

You may need to find:

• moles
• concentration
• solution volume
• mass
• amount of product

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Worked Example: Acid and Base

Consider:

HCl + NaOH → NaCl + H₂O

A student reacts:

100 mL of 0.50 mol/L HCl

with sufficient NaOH.

How many moles of NaOH are required?

Convert volume

100 mL = 0.100 L

Calculate moles HCl

n = cV

n = 0.50 × 0.100

n = 0.050 mol HCl

Use the mole ratio

From the equation:

1 mol HCl : 1 mol NaOH

Therefore:

0.050 mol HCl requires 0.050 mol NaOH

Answer

0.050 mol NaOH

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Finding the Required Volume

Suppose the NaOH in the previous example has a concentration of:

0.25 mol/L

We need:

0.050 mol NaOH

Use:

V = n/c

V = 0.050 / 0.25

V = 0.200 L

Convert:

0.200 L = 200 mL

Answer

200 mL of 0.25 mol/L NaOH

is required.

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A Complete Concentration-to-Volume Problem

Consider:

HCl + NaOH → NaCl + H₂O

What volume of:

0.20 mol/L NaOH

is required to react completely with:

50 mL of 0.40 mol/L HCl?

Convert HCl volume

50 mL = 0.050 L

Calculate HCl moles

n = cV

n = 0.40 × 0.050

n = 0.020 mol HCl

Use the mole ratio

HCl : NaOH = 1 : 1

Therefore:

0.020 mol NaOH

is required.

Calculate NaOH volume

V = n/c

V = 0.020 / 0.20

V = 0.100 L

Convert:

0.100 L = 100 mL

Answer

100 mL NaOH

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When the Mole Ratio Is Not 1:1

Many reactions do not have equal mole ratios.

Consider:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The mole ratio is:

1 mol H₂SO₄ : 2 mol NaOH

This means:

0.10 mol H₂SO₄

requires:

0.20 mol NaOH

Ignoring the coefficients would produce an incorrect answer.

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Worked Example: Sulfuric Acid and Sodium Hydroxide

What volume of:

0.50 mol/L NaOH

is required to react completely with:

100 mL of 0.25 mol/L H₂SO₄?

Equation:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Calculate H₂SO₄ moles

Convert:

100 mL = 0.100 L

Then:

n = cV

n = 0.25 × 0.100

= 0.025 mol H₂SO₄

Apply the mole ratio

H₂SO₄ : NaOH = 1 : 2

Therefore:

0.025 × 2 = 0.050 mol NaOH

Calculate NaOH volume

V = n/c

V = 0.050 / 0.50

= 0.100 L

Convert:

0.100 L = 100 mL

Answer

100 mL NaOH

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Another Non-1:1 Ratio

Consider:

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

The ratio is:

2 mol HCl : 1 mol Ca(OH)₂

Suppose:

200 mL of 0.30 mol/L HCl

react completely.

Calculate HCl moles

200 mL = 0.200 L

n = 0.30 × 0.200

= 0.060 mol HCl

Apply the ratio

2 HCl : 1 Ca(OH)₂

Therefore:

0.060 ÷ 2 = 0.030 mol Ca(OH)₂

If the Ca(OH)₂ solution has concentration:

0.20 mol/L

then:

V = 0.030 / 0.20

= 0.150 L

= 150 mL

Answer

150 mL Ca(OH)₂ solution

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Solution Stoichiometry and Particle Ratios

Balanced equations represent particle relationships as well as mole relationships.

For:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

one formula unit of sulfuric acid reacts with two formula units of sodium hydroxide.

At the mole level:

1 mol H₂SO₄ reacts with 2 mol NaOH

At the particle level, the same proportional relationship applies.

This is why balanced equations are essential.

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Precipitation Reactions

Solution stoichiometry is also important when two aqueous ionic solutions react to form an insoluble solid.

The solid that forms is called a precipitate.

For example:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Silver chloride, AgCl, is insoluble and forms a solid precipitate.

Worked Example: Precipitate Formation

Consider:

AgNO₃ + NaCl → AgCl + NaNO₃

A student mixes:

100 mL of 0.20 mol/L AgNO₃

with excess NaCl.

How many moles of AgCl can form?

Calculate AgNO₃ moles

100 mL = 0.100 L

n = cV

n = 0.20 × 0.100

= 0.020 mol AgNO₃

Apply the mole ratio

AgNO₃ : AgCl = 1 : 1

Therefore:

0.020 mol AgCl

can form.

Answer

0.020 mol AgCl

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Finding the Mass of a Product

We can extend the previous problem.

If:

0.020 mol AgCl

forms, what mass is produced?

Use:

m = nM

Take:

M(AgCl) = 143.5 g/mol

Then:

m = 0.020 × 143.5

m = 2.87 g

Answer

2.87 g AgCl

The full pathway was:

concentration → moles → mole ratio → moles product → mass product

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Multi-Step Example: Product Mass

Consider:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

A student reacts:

250 mL of 0.10 mol/L BaCl₂

with excess sodium sulfate.

Calculate the mass of BaSO₄ formed.

Use:

M(BaSO₄) = 233 g/mol

Calculate BaCl₂ moles

250 mL = 0.250 L

n = 0.10 × 0.250

= 0.025 mol BaCl₂

Apply the mole ratio

BaCl₂ : BaSO₄ = 1 : 1

Therefore:

0.025 mol BaSO₄

forms.

Calculate mass

m = nM

m = 0.025 × 233

= 5.825 g

Answer

Approximately:

5.83 g BaSO₄

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Finding Product Concentration

Sometimes a problem asks for the concentration of a product in the resulting solution.

Consider:

HCl + NaOH → NaCl + H₂O

Suppose:

100 mL of 1.0 mol/L HCl

reacts exactly with:

100 mL of 1.0 mol/L NaOH

Calculate HCl moles

n = 1.0 × 0.100

= 0.100 mol

The 1:1 ratio means:

0.100 mol NaCl

forms.

Assuming the solution volumes are approximately additive:

total volume = 100 + 100 = 200 mL

= 0.200 L

NaCl concentration:

c = n/V

c = 0.100 / 0.200

= 0.50 mol/L

Answer

[NaCl] ≈ 0.50 mol/L

This calculation assumes the final solution volume is approximately the sum of the original solution volumes.

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Reactants in Solution

A chemical equation may include the symbol:

(aq)

This means aqueous.

For example:

HCl(aq)

means hydrogen chloride dissolved in water.

Other state symbols include:

• (s) = solid
• (l) = liquid
• (g) = gas
• (aq) = aqueous

These symbols help us interpret what is physically occurring during a reaction.

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Ions in Aqueous Solutions

Many ionic compounds separate into ions when dissolved.

For example:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

and:

CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq)

This means:

1 mol CaCl₂

produces:

1 mol Ca²⁺

and:

2 mol Cl⁻

So a:

0.50 mol/L CaCl₂

solution contains approximately:

0.50 mol/L Ca²⁺

and:

1.0 mol/L Cl⁻

assuming complete dissociation.

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Worked Example: Ion Concentration

A solution contains:

0.30 mol/L AlCl₃

Assuming complete dissociation:

AlCl₃ → Al³⁺ + 3Cl⁻

For every mole of AlCl₃:

• 1 mol Al³⁺ forms
• 3 mol Cl⁻ form

Therefore:

[Al³⁺] = 0.30 mol/L

and:

[Cl⁻] = 3 × 0.30

= 0.90 mol/L

Answer

Al³⁺ concentration = 0.30 mol/L

Cl⁻ concentration = 0.90 mol/L

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Net Ionic Equations

Some solution reactions can be represented more clearly using net ionic equations.

Consider:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

In solution:

AgNO₃ → Ag⁺ + NO₃⁻

NaCl → Na⁺ + Cl⁻

The ions Na⁺ and NO₃⁻ remain unchanged.

They are called spectator ions.

The particles actually reacting are:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

This is the net ionic equation.

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Spectator Ions

A spectator ion is an ion that is present in the solution but does not participate directly in the chemical change.

In:

Ag⁺ + NO₃⁻ + Na⁺ + Cl⁻ → AgCl + Na⁺ + NO₃⁻

the spectator ions are:

Na⁺

and:

NO₃⁻

They appear unchanged before and after the reaction.

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Neutralization in Solution

Acid-base neutralization is another major application of solution stoichiometry.

For a strong acid and strong base, the central ionic reaction is:

H⁺(aq) + OH⁻(aq) → H₂O(l)

One mole of H⁺ reacts with one mole of OH⁻.

This explains why concentration and volume measurements can be used to determine how much acid or base is required for neutralization.

Solution Stoichiometry and Titration

A titration is an experimental technique used to determine an unknown concentration by reacting a solution with another solution of known concentration.

Typically:

• one solution is measured using a pipette
• another is added from a burette
• an indicator or instrument helps determine when the reaction is complete
• the measured volumes are used in stoichiometric calculations

Solution stoichiometry provides the mathematical foundation for titration calculations.

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Worked Example: Unknown Concentration

Consider:

HCl + NaOH → NaCl + H₂O

A:

25.0 mL

sample of HCl reacts exactly with:

20.0 mL of 0.150 mol/L NaOH

Calculate the concentration of HCl.

Calculate NaOH moles

Convert:

20.0 mL = 0.0200 L

n = cV

n = 0.150 × 0.0200

= 0.00300 mol NaOH

Apply the mole ratio

HCl : NaOH = 1 : 1

Therefore:

n(HCl) = 0.00300 mol

Calculate HCl concentration

Convert:

25.0 mL = 0.0250 L

Use:

c = n/V

c = 0.00300 / 0.0250

= 0.120 mol/L

Answer

HCl concentration = 0.120 mol/L

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Titration with a Different Mole Ratio

Consider:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

A:

25.0 mL

sample of H₂SO₄ reacts with:

30.0 mL of 0.200 mol/L NaOH

Find the H₂SO₄ concentration.

Calculate NaOH moles

30.0 mL = 0.0300 L

n = 0.200 × 0.0300

= 0.00600 mol NaOH

Apply the ratio

H₂SO₄ : NaOH = 1 : 2

Therefore:

n(H₂SO₄) = 0.00600 / 2

= 0.00300 mol

Calculate concentration

25.0 mL = 0.0250 L

c = 0.00300 / 0.0250

= 0.120 mol/L

Answer

H₂SO₄ concentration = 0.120 mol/L

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Limiting Reactants in Solution

When two solutions are mixed, either reactant can be limiting.

Consider:

AgNO₃ + NaCl → AgCl + NaNO₃

Suppose we mix:

100 mL of 0.20 mol/L AgNO₃

and:

100 mL of 0.10 mol/L NaCl

Calculate AgNO₃ moles

n = 0.20 × 0.100

= 0.020 mol

Calculate NaCl moles

n = 0.10 × 0.100

= 0.010 mol

The mole ratio is:

1 : 1

We have:

0.020 mol AgNO₃

but only:

0.010 mol NaCl

Therefore:

NaCl is the limiting reactant.

Only:

0.010 mol AgCl

can form.

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Calculating the Excess Reactant

Continue the previous example.

Initially:

0.020 mol AgNO₃

Only:

0.010 mol

can react because NaCl is limiting.

AgNO₃ remaining:

0.020 − 0.010

= 0.010 mol AgNO₃

Therefore, AgNO₃ is the excess reactant.

If the final solution volume is approximately:

200 mL = 0.200 L

the concentration of remaining AgNO₃ would be:

c = 0.010 / 0.200

= 0.050 mol/L

This type of calculation combines:

• concentration
• moles
• limiting reactants
• excess reactants
• final solution volume

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Producing a Gas from a Solution

Solution stoichiometry can also predict gas production.

Consider:

Mg + 2HCl → MgCl₂ + H₂

Suppose magnesium reacts with:

250 mL of 0.40 mol/L HCl

and magnesium is in excess.

Calculate HCl moles

250 mL = 0.250 L

n = 0.40 × 0.250

= 0.100 mol HCl

Apply the ratio

2 mol HCl : 1 mol H₂

Therefore:

n(H₂) = 0.100 / 2

= 0.050 mol H₂

Answer

0.050 mol H₂

can theoretically form.

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Finding the Mass of a Reactant Required

Consider:

Mg + 2HCl → MgCl₂ + H₂

How much Mg is required to react completely with:

500 mL of 1.0 mol/L HCl?

Use:

M(Mg) = 24.3 g/mol

Calculate HCl moles

500 mL = 0.500 L

n = 1.0 × 0.500

= 0.500 mol HCl

Apply the mole ratio

2 HCl : 1 Mg

Therefore:

n(Mg) = 0.500 / 2

= 0.250 mol Mg

Calculate mass

m = nM

m = 0.250 × 24.3

= 6.075 g

Answer

Approximately:

6.08 g Mg

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Multi-Step Example

Consider:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

A student reacts excess calcium carbonate with:

300 mL of 0.50 mol/L HCl

Calculate the number of moles of CO₂ produced.

Calculate HCl moles

300 mL = 0.300 L

n = 0.50 × 0.300

= 0.150 mol HCl

Apply the mole ratio

2 HCl : 1 CO₂

Therefore:

n(CO₂) = 0.150 / 2

= 0.075 mol CO₂

Answer

0.075 mol CO₂

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A General Problem-Solving Strategy

For solution stoichiometry, use:

BALANCE → MOLES → RATIO → ANSWER

Balance

Write the balanced chemical equation.

Moles

Convert concentration and volume into moles:

n = cV

Ratio

Use the coefficients in the balanced equation.

Answer

Convert the resulting moles into whatever quantity the question asks for:

• concentration
• volume
• mass
• moles
• product quantity

This pathway works for a wide range of solution problems.

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A More Detailed Calculation Map

Depending on the information provided, you might use:

volume + concentration

↓

n = cV

↓

moles of known substance

↓

balanced equation

↓

mole ratio

↓

moles of unknown substance

↓

Then choose:

c = n/V → concentration

V = n/c → volume

m = nM → mass

or leave the answer in moles.

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Common Mistakes

Forgetting to Balance the Equation

Stoichiometric ratios come from the balanced equation.

An unbalanced equation gives incorrect mole relationships.

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Using Volume Directly as Moles

A volume such as:

100 mL

does not tell you the number of moles by itself.

You need concentration:

n = cV

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Forgetting to Convert mL to L

When using:

n = cV

with concentration in mol/L:

250 mL = 0.250 L

not 250 L.

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Ignoring the Mole Ratio

For:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

0.10 mol H₂SO₄ does not react with 0.10 mol NaOH.

It requires:

0.20 mol NaOH

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Using the Wrong Ratio Direction

If:

2HCl : 1Mg

and you know HCl moles:

Mg moles = HCl moles × 1/2

not:

× 2

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Confusing Concentration with Moles

0.50 mol/L

is a concentration.

It is not:

0.50 mol

unless the solution volume is exactly 1.0 L.

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Assuming Equal Volumes Mean Equal Moles

Equal volumes contain equal moles only if their concentrations are also equal.

For example:

100 mL of 1.0 mol/L contains:

0.100 mol

100 mL of 0.20 mol/L contains:

0.020 mol

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Assuming Equal Concentrations Mean Equal Amounts

Two solutions can have the same concentration but different volumes.

Therefore, they can contain different numbers of moles.

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Forgetting the Limiting Reactant

If quantities of both reactants are given:

1. calculate the moles of each
2. compare them using the mole ratio
3. identify the limiting reactant

Do not automatically use the first reactant listed.

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Using Total Volume Too Early

Use each solution's own volume when calculating its initial number of moles.

Only use the combined volume when calculating concentrations after mixing, if the problem allows volumes to be treated as additive.

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Key Terms

Solution stoichiometry — Quantitative calculations involving chemical reactions in solution.

Molar concentration — Number of moles of solute per litre of solution.

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

Aqueous solution — A solution in which water is the solvent.

Precipitate — An insoluble solid formed during a reaction in solution.

Precipitation reaction — A reaction in which dissolved ions combine to form an insoluble solid.

Neutralization — A reaction between an acid and a base.

Titration — A technique in which measured volumes of solutions are reacted to determine an unknown concentration or amount.

Limiting reactant — The reactant consumed first, which limits the amount of product.

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

Spectator ion — An ion that remains unchanged during an ionic reaction.

Net ionic equation — An equation showing only the particles directly involved in the chemical change.

Stoichiometric ratio — The quantitative relationship between substances in a balanced equation.

Equivalence point — The point in a titration at which reactants have been combined in their required stoichiometric proportions.

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Key Takeaways

• Solution stoichiometry combines molar concentration with mole ratios.
• Moles in a solution can be calculated using:

n = cV

• When concentration is in mol/L, volume must be in litres.
• The balanced chemical equation determines the mole ratio.
• A useful pathway is:

concentration + volume → moles → mole ratio → answer

• If the required answer is another solution volume, use:

V = n/c

• If the required answer is concentration, use:

c = n/V

• If the required answer is mass, use:

m = nM

• Acid-base neutralization calculations are common examples of solution stoichiometry.
• Precipitation reactions can be analyzed using concentration and volume data.
• Solution stoichiometry can predict the amount of precipitate or gas formed.
• When quantities of both reactants are provided, the limiting reactant may need to be identified.
• Ionic compounds can produce multiple moles of ions per mole of compound.
• Net ionic equations show the particles actually undergoing chemical change.
• Titration uses solution stoichiometry to determine unknown concentrations.
• Concentration and moles are related but are not the same quantity.

The central strategy is:

BALANCE → MOLES → RATIO → ANSWER

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Check Your Understanding

Concentration and Moles

1. How many moles are present in 500 mL of 0.40 mol/L NaCl?

2. Calculate the moles in 250 mL of 1.2 mol/L HCl.

3. Calculate the moles in 50 mL of 2.0 mol/L NaOH.

4. A solution contains 0.15 mol solute in 300 mL. Calculate its concentration.

5. What volume of 0.50 mol/L solution contains 0.10 mol solute?

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1:1 Reactions

Use:

HCl + NaOH → NaCl + H₂O

6. How many moles of NaOH react with 0.20 mol HCl?

7. How many moles of NaCl form from 0.050 mol HCl?

8. 100 mL of 0.50 mol/L HCl reacts completely. Calculate the moles of NaOH required.

9. If the NaOH in Question 8 has a concentration of 0.25 mol/L, calculate the required volume.

10. Calculate the moles of NaCl formed in Question 8.

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Different Mole Ratios

Use:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

11. How many moles of NaOH react with 0.10 mol H₂SO₄?

12. Calculate the moles of H₂SO₄ in 200 mL of 0.25 mol/L solution.

13. Determine the moles of NaOH required for Question 12.

14. If the NaOH concentration is 0.50 mol/L, calculate the required volume.

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Use:

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

15. 150 mL of 0.40 mol/L HCl reacts completely. Calculate the moles of Ca(OH)₂ required.

16. If the Ca(OH)₂ concentration is 0.20 mol/L, calculate its required volume.

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Precipitation Reactions

Use:

AgNO₃ + NaCl → AgCl + NaNO₃

17. Calculate the moles of AgNO₃ in 250 mL of 0.20 mol/L solution.

18. If NaCl is in excess, calculate the moles of AgCl formed.

19. Calculate the mass of AgCl formed. Use M(AgCl) = 143.5 g/mol.

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Use:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

20. 200 mL of 0.15 mol/L BaCl₂ reacts with excess Na₂SO₄. Calculate the moles of BaSO₄ formed.

21. Calculate the mass of BaSO₄ formed. Use M(BaSO₄) = 233 g/mol.

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Gas-Producing Reactions

Use:

Mg + 2HCl → MgCl₂ + H₂

22. Calculate the moles of HCl in 300 mL of 0.50 mol/L HCl.

23. If Mg is in excess, calculate the moles of H₂ formed.

24. Calculate the moles of Mg required.

25. Calculate the mass of Mg required. Use M(Mg) = 24.3 g/mol.

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Unknown Concentrations

Use:

HCl + NaOH → NaCl + H₂O

26. 25.0 mL HCl reacts exactly with 30.0 mL of 0.100 mol/L NaOH. Calculate the HCl concentration.

27. 20.0 mL NaOH reacts exactly with 25.0 mL of 0.200 mol/L HCl. Calculate the NaOH concentration.

---

Use:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

28. 25.0 mL H₂SO₄ reacts exactly with 40.0 mL of 0.150 mol/L NaOH. Calculate the H₂SO₄ concentration.

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Limiting Reactants

Use:

AgNO₃ + NaCl → AgCl + NaNO₃

29. 100 mL of 0.30 mol/L AgNO₃ is mixed with 200 mL of 0.10 mol/L NaCl. Calculate the moles of each reactant.

30. Identify the limiting reactant.

31. Calculate the moles of AgCl formed.

32. Calculate the moles of excess reactant remaining.

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Ions in Solution

33. A solution contains 0.40 mol/L NaCl. Determine the concentrations of Na⁺ and Cl⁻.

34. A solution contains 0.30 mol/L CaCl₂. Determine the concentrations of Ca²⁺ and Cl⁻.

35. A solution contains 0.20 mol/L AlCl₃. Determine the concentrations of Al³⁺ and Cl⁻.

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Analysis and Application

36. Explain why a balanced equation is essential in solution stoichiometry.

37. Explain why equal volumes of two solutions do not necessarily contain equal numbers of moles.

38. A student uses 50 mL instead of 0.050 L in the equation n = cV when c is expressed in mol/L. Explain the error and its effect on the answer.

39. Describe the complete calculation pathway for determining the mass of a precipitate when the concentration and volume of one reactant are given and the other reactant is in excess.

40. Two aqueous reactants are mixed and the concentration and volume of both are known. Describe how you would determine the limiting reactant, the amount of product formed, and the quantity of excess reactant remaining.