Laboratory Applications

Stoichiometry is essential in laboratory chemistry because it allows us to calculate how much of each substance is required for a reaction and how much product should theoretically be produced.

Before carrying out an experiment, stoichiometry can answer questions such as:

• How much reactant should be weighed?
• What volume of solution is required?
• Which reactant will be limiting?
• How much product should form?
• Is one reactant being used in excess?
• How does the experimental result compare with the theoretical result?

After an experiment, the same calculations help us evaluate the quality of our results.

A useful overall pathway is:

PLAN → MEASURE → REACT → CALCULATE → COMPARE → EVALUATE

Stoichiometry Before an Experiment

Laboratory investigations should usually be planned quantitatively before chemicals are mixed.

Suppose the reaction is:

Mg + 2HCl → MgCl₂ + H₂

The balanced equation tells us:

1 mol Mg : 2 mol HCl : 1 mol MgCl₂ : 1 mol H₂

If we know how much magnesium will be used, we can calculate:

• the amount of HCl required
• the amount of MgCl₂ expected
• the amount of H₂ expected

This helps prevent unnecessary use of chemicals.

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The Basic Laboratory Calculation

Many laboratory calculations follow:

MASS → MOLES → MOLE RATIO → MOLES → MASS

Use:

n = m/M

where:

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

To find mass:

m = nM

The balanced chemical equation provides the mole ratio between substances.

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Worked Example: Magnesium and Hydrochloric Acid

Reaction:

Mg + 2HCl → MgCl₂ + H₂

Suppose:

2.4 g Mg

is reacted with excess hydrochloric acid.

Use:

M(Mg) = 24 g/mol

Calculate moles of Mg:

n = 2.4 / 24

= 0.10 mol

The equation shows:

1 mol Mg → 1 mol H₂

Therefore:

0.10 mol Mg → 0.10 mol H₂

It also shows:

1 mol Mg → 1 mol MgCl₂

Therefore:

0.10 mol MgCl₂

should theoretically form.

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Calculating Required Reactant Quantities

The balanced equation can be used to determine exactly how much of another reactant is needed.

For:

Mg + 2HCl → MgCl₂ + H₂

we calculated:

0.10 mol Mg

The equation requires:

1 mol Mg : 2 mol HCl

Therefore:

0.10 mol Mg requires 0.20 mol HCl

If the HCl solution has a concentration of:

1.0 mol/L

use:

n = cV

Rearrange:

V = n/c

Therefore:

V = 0.20 / 1.0

= 0.20 L

Convert:

0.20 L = 200 mL

Answer

200 mL of 1.0 mol/L HCl

is theoretically required.

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Using Solution Concentration

Many laboratory reactions involve solutions rather than solid reactants.

Use:

n = cV

where:

• n = moles
• c = concentration in mol/L
• V = volume in L

Remember:

1000 mL = 1 L

Therefore:

25.0 mL = 0.0250 L

This conversion is essential.

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Worked Example: Solution Reaction

Consider:

NaOH + HCl → NaCl + H₂O

Suppose:

25.0 mL of 0.200 mol/L NaOH

is used.

Calculate the amount of HCl required.

Find moles of NaOH

Convert volume:

25.0 mL = 0.0250 L

Use:

n = cV

n = 0.200 × 0.0250

= 0.00500 mol

The mole ratio is:

1 mol NaOH : 1 mol HCl

Therefore:

n(HCl) = 0.00500 mol

If the HCl concentration is also:

0.200 mol/L

then:

V = n/c

V = 0.00500 / 0.200

= 0.0250 L

= 25.0 mL

Answer

25.0 mL HCl

is required.

Stoichiometry and Titration

A titration uses a solution of known concentration to determine the amount or concentration of another substance.

For a simple reaction:

HCl + NaOH → NaCl + H₂O

the ratio is:

1 : 1

If a titration shows that:

0.0040 mol NaOH

reacted completely, then:

0.0040 mol HCl

must also have reacted.

For reactions with different coefficients, the appropriate mole ratio must be used.

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Titration with a Different Mole Ratio

Consider:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The mole ratio is:

1 mol H₂SO₄ : 2 mol NaOH

Suppose a titration uses:

0.010 mol H₂SO₄

Then:

0.020 mol NaOH

is required.

A common mistake would be to assume a 1:1 ratio.

The balanced equation must always be checked first.

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Limiting Reactants in Laboratory Experiments

When two reactants are mixed, one may be completely consumed before the other.

This is the limiting reactant.

The limiting reactant determines the maximum possible amount of product.

Consider:

2H₂ + O₂ → 2H₂O

Suppose we have:

4 mol H₂

and:

3 mol O₂

To react with 4 mol H₂, we need:

2 mol O₂

But 3 mol O₂ is available.

Therefore:

H₂ is limiting

and:

O₂ is in excess

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Why Laboratories Sometimes Use an Excess Reactant

Experiments sometimes deliberately use one reactant in excess.

This can help ensure that another reactant:

• reacts completely
• determines the product amount
• can be studied more easily

For example, if the purpose of an experiment is to determine how much product forms from a known mass of magnesium, hydrochloric acid may be supplied in excess.

Then magnesium is the limiting reactant.

This makes the calculation simpler because all the magnesium should react.

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Worked Example: Limiting Reactant

Consider:

2Mg + O₂ → 2MgO

Suppose:

0.30 mol Mg

and:

0.10 mol O₂

are available.

The equation requires:

2 mol Mg : 1 mol O₂

To react with 0.30 mol Mg:

0.30 × 1/2 = 0.15 mol O₂

But only:

0.10 mol O₂

is available.

Therefore:

O₂ is the limiting reactant.

From:

1 mol O₂ → 2 mol MgO

we get:

0.10 mol O₂ → 0.20 mol MgO

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Theoretical Yield

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

It assumes:

• complete reaction
• no product is lost
• no side reactions occur
• measurements are accurate
• reactants have the assumed purity

Suppose stoichiometry predicts:

5.00 g product

Then:

theoretical yield = 5.00 g

The theoretical yield is calculated rather than measured.

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Actual Yield

The actual yield is the amount of product actually obtained during the experiment.

Suppose:

theoretical yield = 5.00 g

but after filtering, drying, and weighing, the student obtains:

4.20 g

Then:

actual yield = 4.20 g

The difference between these values can provide useful information about the experiment.

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Percentage Yield

Percentage yield compares the actual yield with the theoretical yield.

Use:

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

For:

actual yield = 4.20 g

and:

theoretical yield = 5.00 g

calculate:

percentage yield = (4.20 / 5.00) × 100

= 84.0%

Answer

Percentage yield = 84.0%

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Worked Example: Theoretical and Percentage Yield

Consider:

2Mg + O₂ → 2MgO

A student reacts:

2.40 g Mg

with excess oxygen.

Use:

M(Mg) = 24.0 g/mol

M(MgO) = 40.0 g/mol

Calculate moles Mg

n = 2.40 / 24.0

= 0.100 mol

The ratio is:

1 mol Mg : 1 mol MgO

Therefore:

0.100 mol MgO

should form.

Calculate theoretical mass

m = nM

m = 0.100 × 40.0

= 4.00 g

Suppose the student actually obtains:

3.60 g MgO

Calculate percentage yield:

(3.60 / 4.00) × 100

= 90.0%

Results

Theoretical yield = 4.00 g

Actual yield = 3.60 g

Percentage yield = 90.0%

Why Actual Yield May Be Lower

A percentage yield below 100% is common.

Possible reasons include:

Incomplete reaction

Some reactant may not have reacted.

Product lost during transfer

Some material may remain:

• on glassware
• on filter paper
• on stirring rods
• in transfer containers

Product lost during filtration

Small amounts may pass through the filter or remain in the apparatus.

Product lost during heating

Material may spit, splash, or escape.

Side reactions

Reactants may form unwanted products.

Reversible reactions

The reaction may not proceed completely toward products.

Product remains dissolved

Some product may stay in solution rather than being recovered.

These explanations should be linked to the actual experimental procedure whenever possible.

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Can Percentage Yield Be Greater Than 100%?

Mathematically, an experimental calculation can produce a result above 100%.

For example:

Theoretical yield:

5.00 g

Measured product:

5.40 g

Percentage yield:

(5.40 / 5.00) × 100

= 108%

This does not mean the reaction created more product than theoretically possible.

Instead, the measured "product" probably contains additional mass.

Possible causes include:

• product was still wet
• contamination
• unreacted reactant remained
• another substance was weighed with the product
• weighing error

A yield above 100% is therefore evidence that the experimental result should be investigated.

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Wet Products and Percentage Yield

Suppose a precipitate should have a theoretical mass of:

2.50 g

A student weighs:

2.80 g

before the sample is completely dry.

Calculated yield:

(2.80 / 2.50) × 100

= 112%

The most likely problem is that the measured mass includes:

product + water

After proper drying, the measured mass should decrease.

This is why many laboratory procedures specify:

dry to constant mass

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Constant Mass

A sample is at constant mass when repeated cycles of heating, cooling, and weighing produce approximately the same mass.

For example:

The final repeated value suggests:

constant mass ≈ 4.15 g

Constant mass provides evidence that processes such as drying or reaction have reached completion.

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Interpreting Experimental Data

Stoichiometry can help determine whether experimental data are reasonable.

Suppose the theoretical product mass is:

6.50 g

Three groups obtain:

• Group A: 6.20 g
• Group B: 4.10 g
• Group C: 7.30 g

Group A is reasonably close to the theoretical value.

Group B may have:

• lost product
• had an incomplete reaction
• made a transfer error

Group C produced an apparent yield above 100%, suggesting:

• wet product
• contamination
• incorrect weighing
• another experimental problem

Stoichiometry provides a benchmark against which experimental results can be evaluated.

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Percentage Error and Percentage Yield Are Different

Do not confuse percentage yield with percentage error.

Percentage yield compares product obtained with the theoretical amount:

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

Percentage error compares an experimental measurement with an accepted value:

percentage error = |experimental − accepted| / accepted × 100

They answer different questions.

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Precipitation Reactions

Stoichiometry is often used in experiments where an insoluble solid forms.

This solid is called a precipitate.

For example:

AgNO₃ + NaCl → AgCl + NaNO₃

AgCl is insoluble and forms a precipitate.

The mole ratio is:

1 mol AgNO₃ : 1 mol NaCl : 1 mol AgCl

If:

0.010 mol AgNO₃

reacts with excess NaCl, theoretically:

0.010 mol AgCl

forms.

Using:

M(AgCl) ≈ 143.5 g/mol

mass:

m = 0.010 × 143.5

= 1.435 g

Therefore:

theoretical AgCl yield ≈ 1.44 g

Filtering and Drying a Precipitate

A typical precipitation investigation may involve:

1. measuring reactant solutions
2. mixing the solutions
3. allowing the precipitate to form
4. filtering the mixture
5. washing the precipitate
6. drying the precipitate
7. measuring its mass
8. comparing the actual mass with the theoretical mass

Stoichiometry therefore connects directly to laboratory technique.

Poor technique can affect the measured yield.

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Gas-Producing Reactions

Stoichiometry can also predict quantities of gases.

Consider:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

The equation shows:

1 mol CaCO₃ → 1 mol CO₂

Therefore, if:

0.050 mol CaCO₃

reacts completely:

0.050 mol CO₂

should form.

If gas volume relationships are known for the experimental conditions, the expected volume can also be calculated.

Measuring Gas in the Laboratory

Gas production may be measured using:

• a gas syringe
• water displacement
• pressure measurements
• mass loss

Experimental gas volumes may differ from theoretical predictions because of:

• leaks
• gas dissolving in water
• incomplete reactions
• temperature differences
• pressure differences
• measurement uncertainty

These factors should be considered when evaluating experimental results.

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Using Mass Loss

Consider:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

If the CO₂ escapes from an open reaction vessel, the mass of the apparatus may decrease.

Suppose:

Initial mass:

152.40 g

Final mass:

150.20 g

Mass lost:

152.40 − 150.20

= 2.20 g

If CO₂ is the only substance leaving the system, then:

mass CO₂ = 2.20 g

Molar mass of CO₂:

44.0 g/mol

Moles CO₂:

2.20 / 44.0

= 0.0500 mol

Therefore:

0.0500 mol CaCO₃

must have reacted.

Stoichiometry allows an indirect measurement to reveal the amount of reactant consumed.

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Planning a Precipitation Experiment

Suppose you want to produce approximately:

2.87 g AgCl

Reaction:

AgNO₃ + NaCl → AgCl + NaNO₃

Molar mass:

AgCl = 143.5 g/mol

Calculate required moles:

n = 2.87 / 143.5

= 0.0200 mol

The ratio is:

1 AgNO₃ : 1 AgCl

Therefore:

0.0200 mol AgNO₃

is theoretically required.

If using:

0.500 mol/L AgNO₃

then:

V = n/c

= 0.0200 / 0.500

= 0.0400 L

= 40.0 mL

This calculation can be completed before entering the laboratory.

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Choosing Suitable Quantities

Laboratory stoichiometry is also important for safety and practicality.

An experiment should use enough material to produce measurable results, but not unnecessarily large quantities.

Smaller-scale experiments can:

• reduce chemical waste
• reduce costs
• reduce exposure to hazardous substances
• make cleanup easier
• reduce environmental impact

Stoichiometry helps determine the minimum practical quantities needed.

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Stoichiometry and Experimental Design

Suppose students want to investigate the percentage yield of magnesium oxide.

They need to decide:

• mass of magnesium
• amount of oxygen available
• suitable crucible size
• expected product mass
• balance precision
• number of trials

If the expected product mass is extremely small compared with the balance uncertainty, the experiment may produce poor-quality data.

Therefore, stoichiometry can help design experiments as well as analyze them.

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Measurement Uncertainty

Every laboratory measurement has some uncertainty.

Suppose a balance reads to:

±0.01 g

Measuring:

0.10 g

of substance has a relatively large percentage uncertainty.

Measuring:

5.00 g

with the same balance has a much smaller relative uncertainty.

Therefore, the quantities chosen for an experiment can influence the reliability of stoichiometric results.

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Evaluating Results

A strong evaluation does more than say:

"There was human error."

Instead, identify a specific problem and explain how it affected the result.

For example:

Some MgO powder escaped from the crucible during heating. This reduced the measured actual yield, causing the calculated percentage yield to be lower than the true value.

This identifies:

1. the specific error
2. what measurement it affected
3. the direction of the effect
4. the consequence for the final calculation

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Error Direction

Understanding whether an error makes a result too high or too low is particularly important.

This type of reasoning helps connect stoichiometry with experimental evaluation.

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Worked Practical Example

A student investigates:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

The student uses:

5.00 g CaCO₃

with excess HCl.

Use:

M(CaCO₃) = 100 g/mol

M(CO₂) = 44 g/mol

Calculate moles CaCO₃

n = 5.00 / 100

= 0.0500 mol

Use the mole ratio

1 mol CaCO₃ → 1 mol CO₂

Therefore:

0.0500 mol CO₂

Calculate theoretical CO₂ mass

m = 0.0500 × 44

= 2.20 g

Suppose the measured mass loss is:

2.02 g

Calculate percentage yield

percentage yield = (2.02 / 2.20) × 100

≈ 91.8%

Possible reasons for the lower result include:

• incomplete reaction
• CO₂ remaining dissolved
• measurement uncertainty

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Multi-Step Laboratory Example

Consider:

2Al + 3CuCl₂ → 2AlCl₃ + 3Cu

A student reacts:

0.540 g Al

with excess CuCl₂.

Use:

M(Al) = 27.0 g/mol

M(Cu) = 63.5 g/mol

Find moles Al

n = 0.540 / 27.0

= 0.0200 mol

Apply the mole ratio

From:

2 mol Al → 3 mol Cu

therefore:

0.0200 × (3/2)

= 0.0300 mol Cu

Find theoretical copper mass

m = 0.0300 × 63.5

= 1.905 g

Approximately:

1.91 g Cu

Suppose the student collects:

1.72 g Cu

Percentage yield:

(1.72 / 1.905) × 100

≈ 90.3%

Results

Theoretical yield ≈ 1.91 g Cu

Actual yield = 1.72 g Cu

Percentage yield ≈ 90.3%

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Using Experimental Results to Find an Unknown

Stoichiometry can sometimes be used to determine an unknown quantity.

Suppose an unknown mass of Mg reacts completely:

Mg + 2HCl → MgCl₂ + H₂

The experiment produces:

0.050 mol H₂

The mole ratio is:

1 mol Mg : 1 mol H₂

Therefore:

n(Mg) = 0.050 mol

Using:

M(Mg) = 24 g/mol

m = 0.050 × 24

= 1.20 g

Therefore, the original magnesium mass was:

1.20 g

Stoichiometry can therefore work both forwards and backwards.

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Comparing Trials

Suppose the theoretical yield is:

4.00 g

Three trials produce:

These results are relatively close to one another.

This suggests good repeatability.

The mean actual yield is:

(3.80 + 3.76 + 3.82) / 3

= 3.79 g

Mean percentage yield:

(3.79 / 4.00) × 100

≈ 94.8%

Repeated trials provide stronger evidence than a single measurement.

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Precision and Accuracy

These ideas are useful when evaluating laboratory stoichiometry.

Precision describes how closely repeated measurements agree with one another.

Accuracy describes how close a measurement is to the accepted or expected value.

A set of trials could be:

• precise and accurate
• precise but inaccurate
• imprecise but approximately accurate on average
• neither precise nor accurate

Stoichiometric theoretical values can provide a useful reference when assessing experimental accuracy.

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Improving a Laboratory Investigation

Possible improvements include:

• using more precise measuring equipment
• ensuring complete reaction
• preventing loss during transfer
• using clean, dry apparatus
• drying products to constant mass
• preventing gas leaks
• using an appropriate excess reactant
• repeating the experiment
• calculating and controlling reactant quantities beforehand

Improvements should address a specific weakness rather than simply saying "be more careful."

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Laboratory Safety and Stoichiometry

Stoichiometric planning can also support safer laboratory work.

Using appropriate quantities helps avoid:

• unnecessary chemical exposure
• excessive gas production
• excessive heat generation
• overflowing reaction vessels
• unnecessary hazardous waste

However, stoichiometric calculations do not replace a proper risk assessment.

Hazards must still be identified and appropriate safety procedures followed.

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Common Mistakes

Forgetting to Balance the Equation

Stoichiometric ratios come from the balanced equation.

Never calculate mole ratios from an unbalanced equation.

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Using Mass Ratios as Mole Ratios

Coefficients represent moles, not grams.

For:

2Mg + O₂ → 2MgO

the ratio is:

2 mol : 1 mol : 2 mol

not:

2 g : 1 g : 2 g

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Forgetting to Convert mL to L

When using:

n = cV

volume must usually be in litres.

For example:

25 mL = 0.025 L

not:

25 L

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Ignoring the Limiting Reactant

When amounts of two reactants are provided, determine which one limits product formation.

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Using Actual Yield to Calculate Theoretical Yield Directly

The theoretical yield comes from:

• reactant quantity
• balanced equation
• stoichiometric ratio

The actual yield comes from the experiment.

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Reversing Percentage Yield

Correct:

percentage yield = (actual / theoretical) × 100

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Assuming a Yield Above 100% Is Excellent

A yield above 100% usually indicates an experimental problem such as:

• wet product
• contamination
• unreacted material

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Giving Vague Error Explanations

Instead of:

"human error"

write something specific, such as:

"Some precipitate remained on the stirring rod during transfer, decreasing the measured actual yield."

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Ignoring Error Direction

Always consider whether an error would make the result:

too high

or:

too low

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Key Terms

Stoichiometry — The quantitative relationship between reactants and products in chemical reactions.

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

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

Limiting reactant — The reactant consumed first, which determines the maximum product quantity.

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

Theoretical yield — The maximum product quantity predicted by stoichiometry.

Actual yield — The amount of product actually obtained experimentally.

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

Titration — A laboratory technique using a solution of known concentration to determine the amount or concentration of another substance.

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

Constant mass — A mass that remains essentially unchanged after repeated drying or heating and weighing.

Experimental error — A factor that causes an experimental measurement to differ from its expected or accepted value.

Measurement uncertainty — The range within which a measured value is reasonably expected to lie.

Precision — The closeness of repeated measurements to one another.

Accuracy — The closeness of an experimental result to an accepted or expected value.

Repeatability — The degree to which repeated measurements under the same conditions produce similar results.

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Key Takeaways

• Stoichiometry is used both before and after laboratory experiments.
• Before an experiment, it helps determine suitable reactant quantities.
• After an experiment, it helps evaluate the results.
• Always begin with a balanced chemical equation.
• Coefficients provide mole ratios.
• For mass problems, a common pathway is:

MASS → MOLES → MOLE RATIO → MOLES → MASS

• For solutions:

n = cV

• Convert mL to L when necessary.
• The limiting reactant determines the maximum amount of product.
• A reactant may deliberately be placed in excess to ensure another reactant reacts completely.
• The theoretical yield is calculated from stoichiometry.
• The actual yield is measured experimentally.
• Percentage yield is:

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

• Yields below 100% may result from incomplete reactions or product losses.
• Apparent yields above 100% usually indicate wet, contaminated, or impure product.
• Stoichiometry can be used to interpret titrations, precipitation reactions, gas-producing reactions, and mass-change experiments.
• Experimental errors should be linked to their effect on the final result.
• Repeated trials can improve confidence in experimental conclusions.
• Stoichiometric planning can reduce chemical waste, cost, and unnecessary hazards.

The overall laboratory pathway is:

PLAN → CALCULATE → MEASURE → REACT → COMPARE → EVALUATE

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Check Your Understanding

Basic Concepts

1. Explain why stoichiometry is useful before beginning a laboratory reaction.

2. What information does a balanced chemical equation provide for stoichiometric calculations?

3. Define a limiting reactant.

4. Explain why a laboratory experiment might deliberately use one reactant in excess.

5. Distinguish between theoretical yield and actual yield.

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Mass Calculations

Use:

Mg + 2HCl → MgCl₂ + H₂

6. Calculate the moles of Mg in 4.80 g Mg. Use M(Mg) = 24.0 g/mol.

7. Determine the moles of HCl required to react completely with the Mg in Question 6.

8. Determine the theoretical moles of H₂ produced.

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Use:

2Mg + O₂ → 2MgO

9. Calculate the theoretical mass of MgO formed from 1.20 g Mg.

10. Calculate the theoretical mass of MgO formed from 3.60 g Mg.

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Solutions

11. Calculate the number of moles in 25.0 mL of 0.400 mol/L NaOH.

12. For HCl + NaOH → NaCl + H₂O, determine the moles of HCl required to react with the NaOH in Question 11.

13. If the HCl concentration is 0.200 mol/L, calculate the required volume.

14. Explain why 25 mL cannot simply be entered as V = 25 when using n = cV with concentration in mol/L.

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Limiting Reactants

Use:

2H₂ + O₂ → 2H₂O

15. A reaction contains 6 mol H₂ and 2 mol O₂. Identify the limiting reactant.

16. Calculate the maximum moles of H₂O produced.

17. Determine the amount of excess reactant remaining.

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Percentage Yield

18. A reaction has a theoretical yield of 8.00 g and an actual yield of 6.80 g. Calculate the percentage yield.

19. A reaction theoretically produces 12.5 g product but produces only 10.0 g. Calculate the percentage yield.

20. A theoretical yield is 5.00 g and the experimental mass is 5.40 g. Calculate the apparent percentage yield.

21. Suggest two reasons for the result in Question 20.

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Experimental Analysis

22. A student loses some solid product while transferring it from a beaker to filter paper. Would the calculated percentage yield be too high or too low? Explain.

23. A student weighs a precipitate while it is still wet. Would the percentage yield be too high or too low? Explain.

24. A reaction does not go to completion. How would this probably affect percentage yield?

25. Explain why drying a product to constant mass improves the reliability of a yield calculation.

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Precipitation

Use:

AgNO₃ + NaCl → AgCl + NaNO₃

26. If 0.0200 mol AgNO₃ reacts with excess NaCl, calculate the theoretical moles of AgCl.

27. Using M(AgCl) = 143.5 g/mol, calculate the theoretical mass of AgCl.

28. If 2.60 g AgCl is collected, calculate the percentage yield.

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Gas-Producing Reactions

Use:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

29. Calculate the moles of CaCO₃ in 10.0 g. Use M(CaCO₃) = 100 g/mol.

30. Determine the theoretical moles of CO₂ produced.

31. Determine the theoretical mass of CO₂. Use M(CO₂) = 44 g/mol.

32. The experiment loses only 3.90 g of mass. Calculate the percentage yield based on the expected CO₂ mass.

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Practical Application

33. A student needs 0.0250 mol NaOH for an experiment. Calculate the volume of 0.500 mol/L NaOH required.

34. A reaction should theoretically produce 4.50 g precipitate. Three trials produce 4.20 g, 4.18 g, and 4.22 g. Comment on the repeatability of the results.

35. Calculate the mean actual yield for Question 34.

36. Calculate the percentage yield using the mean result.

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Evaluation

37. Explain why saying only "human error" is a weak evaluation of an experiment.

38. A student obtains a 115% yield of a solid product. Give three specific experimental causes that could explain the result.

39. Explain how stoichiometric calculations can help make a laboratory investigation safer and reduce waste.

40. Describe how you would use stoichiometry to plan, carry out, analyze, and evaluate a laboratory investigation from the balanced chemical equation through to the final percentage yield.