Limiting Reactants & Percent Yield — Print · Grade 11 Science

Chapter: Stoichiometry & Yield

Limiting Reactants & Percent Yield

Learning goals

Readings for this unit

Before this unit

Stoichiometry and percentage ratios.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why is the reactant with less starting mass not automatically limiting?

Illustrative evidence case. Two reactants have different molar masses and different coefficients in the recipe.

Compare product-making capacity, not whichever reactant looks smallerCompare product-making capacity, not whichever reactant looks smaller5 H₂ entities + 2 O₂ entities2H₂ + O₂ → 2H₂OCapacity: H₂ could make 5 H₂O; O₂ could make 4 H₂O.HHHHHHHHHHOOOO
Original selected count model for complete reaction; it omits mechanism, energy and real molecular scale.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why is the reactant with less starting mass not automatically limiting?

Your first model

Convert each starting amount into its capacity to form the same product.

Build the language from the idea

limiting reactant
The reactant that runs out first, capping the amount of product a reaction can make.
excess reactant
A reactant that is left over after the limiting reactant is completely used up.
theoretical yield
The maximum amount of product predicted by stoichiometry, assuming the reaction goes perfectly and the limiting reactant is fully converted.
actual yield
The amount of product actually collected when the reaction is run in the real world.
percent yield
The actual yield expressed as a percentage of the theoretical yield: (actual ÷ theoretical) × 100%.

Real reactions rarely start with a perfect, exactly balanced set of ingredients. Usually you have a bit more of one reactant than another, and that imbalance decides everything. Think about building bicycles from frames and wheels: each bike needs one frame and two wheels. If you have 4 frames but only 6 wheels, you can build just 3 bikes — the wheels run out first, and a leftover frame sits uselessly on the bench. Chemistry works exactly the same way, and learning to spot which ingredient runs out is the key to predicting a real reaction's output.

The reactant that runs out first is the limiting reactant, because it limits how much product can form. Whatever is left over when it is gone is the excess reactant. Here is the essential insight: the limiting reactant, not the total amount of stuff, sets the ceiling on the product. Pour in ten times the excess reactant and it changes nothing — once the limiting reactant is spent, the reaction simply stops. Finding the limiting reactant is therefore the first move in any realistic yield calculation.

To identify it, you let each reactant 'vote' on how much product it could make. Convert each reactant's amount to moles, then use the mole ratio to calculate how much product each one alone would produce. The reactant that predicts the smaller amount of product is the limiting reactant, and that smaller number is the amount that actually forms. Take the ammonia synthesis N₂ + 3H₂ → 2NH₃ with 2.0 mol of N₂ and 3.0 mol of H₂. Nitrogen could make 2.0 × 2 = 4.0 mol of NH₃, but hydrogen could make only 3.0 × (2 ÷ 3) = 2.0 mol. Hydrogen predicts less, so hydrogen is limiting and just 2.0 mol of ammonia forms — the leftover nitrogen is excess.

Compare product-making capacity, not whichever reactant looks smallerCompare product-making capacity, not whichever reactant looks smaller5 H₂ entities + 2 O₂ entities2H₂ + O₂ → 2H₂OCapacity: H₂ could make 5 H₂O; O₂ could make 4 H₂O.HHHHHHHHHHOOOO
Original selected count model for complete reaction; it omits mechanism, energy and real molecular scale.

Even after you have found the limiting reactant, the amount you calculate is still an idealized ceiling. Stoichiometry gives you the theoretical yield: the maximum product possible if every bit of the limiting reactant converts perfectly and nothing is lost. It is the reaction's best-case promise on paper. But laboratories are messier than paper, and the real world almost never delivers the full theoretical amount.

Leftover is initial amount minus the amount consumed by the limiting extentLeftover is initial amount minus the amount consumed by the limiting extentN₂ + 3H₂ → 2NH₃InitialUsedLeftN₂2.00 mol1.00 mol1.00 molH₂3.00 mol3.00 mol0NH₃0formed 2.00mol2.00 molUse the limiting amount to calculate consumption of every partner.
Original theoretical inventory; actual ammonia formation involves equilibrium and kinetics not calculated by this table.

What you actually scrape out of the flask at the end is the actual yield, and it is nearly always less than the theoretical yield. Product gets left clinging to glassware, some reactant wanders off into unwanted side reactions, impurities creep in, and some reactions simply never run fully to completion. To measure how well a reaction performed, chemists compare the two with the percent yield: actual yield divided by theoretical yield, times 100%. If a reaction promised 40.0 g of product but you collected only 34.0 g, the percent yield is (34.0 ÷ 40.0) × 100% = 85%.

Percent yield compares measured target recovery with a theoretical referencePercent yield compares measured target recovery with a theoretical referenceMass (g)Theoretical product10Recovered target8.2Percent yield = 8.2/10 ×100 = 82%Wet or impure recovered material can inflate the apparent yield.
Original selected product masses; recovered target identity and dryness are assumptions to be evaluated, not supplied assay results.

Percent yield is the reality check that turns tidy stoichiometry into practical chemistry — and it carries real economic weight. A pharmaceutical company synthesizing an expensive drug watches percent yield obsessively, because every percentage point lost is wasted money and raw material. Chemists redesign reactions, purify more carefully, and choose conditions specifically to push that number higher. Together, limiting reactants and percent yield answer the unit's essential question with honest precision: stoichiometry tells you the most a reaction could make, and percent yield tells you how much of that promise it actually kept.

Model check: The smaller initial mass is not necessarily the limiting reactant. H₂ limits; it supports 2molNH₃ and consumes 1molN₂, leaving 1molN₂. Compare potential products, not raw amounts.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Compare product capacity rather than starting mass

A limiting reactant is exhausted first in the stated complete-reaction model and therefore sets the maximum target product. Compare each starting amount divided by its coefficient, or convert each into possible target amount. Five H₂ entities and two O₂ entities can form four H₂O entities, leaving one H₂. The smaller-looking count alone is not the rule: the balanced recipe determines the demand. Real amounts usually contain enormous particle counts, while the drawing isolates the integer-ratio idea.

Compare product-making capacity, not whichever reactant looks smallerCompare product-making capacity, not whichever reactant looks smaller5 H₂ entities + 2 O₂ entities2H₂ + O₂ → 2H₂OCapacity: H₂ could make 5 H₂O; O₂ could make 4 H₂O.HHHHHHHHHHOOOO
Original selected count model for complete reaction; it omits mechanism, energy and real molecular scale.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Which reactant limits and how much H₂ remains in the selected inventory?

Visual reasoning — after your attempt

O₂ limits: two O₂ use four H₂ to form four water entities, leaving one H₂ entity.

The limiting reactant supports the smallest possible target-product amount under the balanced model. Convert each reactant to moles and calculate its product capacity; the smallest gram mass is not automatically limiting. Excess means some reactant remains if the modeled reaction completes.

A light sample of one substance may contain more moles than a heavier sample of another. The coefficient requirement also matters, so compare n/coefficient or equivalent product capacities.

Pause and explain

Why is smaller reactant mass not a reliable limiting rule?

Check your reasoning after answering

Molar masses and stoichiometric coefficients differ. The limiting decision must compare usable amounts relative to the reaction ratio.

Part 2 · Calculate leftover using the limiting amount

Once the limiting amount is known, use its reaction extent to calculate each partner’s consumption. In the selected nitrogen/hydrogen inventory, three moles of hydrogen consume one mole of nitrogen and form two moles of ammonia; the remaining nitrogen is one mole. A negative leftover signals inconsistent arithmetic or choosing the wrong limiting extent. Keep theoretical consumption distinct from a measured equilibrium mixture, where incomplete conversion can leave both reactants present.

Leftover is initial amount minus the amount consumed by the limiting extentLeftover is initial amount minus the amount consumed by the limiting extentN₂ + 3H₂ → 2NH₃InitialUsedLeftN₂2.00 mol1.00 mol1.00 molH₂3.00 mol3.00 mol0NH₃0formed 2.00mol2.00 molUse the limiting amount to calculate consumption of every partner.
Original theoretical inventory; actual ammonia formation involves equilibrium and kinetics not calculated by this table.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why does starting with 2.00 mol N₂ not guarantee production of 4.00 mol NH₃ here?

Visual reasoning — after your attempt

Only 3.00 mol H₂ is available, enough to react with 1.00 mol N₂ and form 2.00 mol NH₃ under the complete-reaction model.

Once the limiting amount is identified, calculate the amount of each other reactant consumed from the balanced ratio. Subtract from its initial amount to find leftover. A negative leftover is a warning that the limiting choice or ratio was wrong.

For N₂+3H₂, 0.5 mol N₂ uses 1.5 mol H₂. If hydrogen starts at 2.0 mol, 0.5 mol remains under the complete-conversion model.

Pause and explain

What check should an excess-reactant calculation pass?

Check your reasoning after answering

The consumed amount cannot exceed the available amount, and the leftover should be nonnegative within the model and measurement uncertainty.

Part 3 · Distinguish theoretical yield and measured recovery

Percent yield is actual recovered target amount divided by the theoretical amount times 100%. The comparison must use compatible quantities and product identity. Transfer losses, incomplete conversion and competing reactions can lower recovery, while water or contamination can inflate a gross recovered mass. A value above 100% usually indicates a measurement, purity or model problem rather than creation of extra atoms. A single yield number cannot distinguish all causes; examine the product assay and procedure.

Percent yield compares measured target recovery with a theoretical referencePercent yield compares measured target recovery with a theoretical referenceMass (g)Theoretical product10Recovered target8.2Percent yield = 8.2/10 ×100 = 82%Wet or impure recovered material can inflate the apparent yield.
Original selected product masses; recovered target identity and dryness are assumptions to be evaluated, not supplied assay results.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Does an 82% isolated yield prove that exactly 18% of the reactants never reacted?

Visual reasoning — after your attempt

No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.

Theoretical yield is predicted from the limiting amount. Percent yield=actual target yield/theoretical yield×100%. Both must refer to the same product and compatible units. Product contamination or residual solvent can inflate mass without increasing the amount of pure target substance.

A reported yield over 100% calls for examining purity, drying, measurements, or assumptions. It is not evidence that atom conservation failed.

Pause and explain

Does 110% recovered mass prove more pure product was formed than theoretically possible?

Check your reasoning after answering

No. Contamination, solvent, errors, or an incomplete theoretical model can inflate the measured result.

Part 4 · Separate conversion from selectivity

Conversion describes how much starting reactant is consumed. Selectivity describes distribution toward a target among possible products under a defined basis. Isolated yield also includes how much target is recovered. A process can have high conversion but poor target selectivity, or good target formation with losses during purification. These terms require explicit denominators and reaction stoichiometry. Use a material-flow account rather than multiply vague percentages that were measured on incompatible bases.

A reaction can consume material without delivering all of it as isolated targetA reaction can consume material without delivering all of it as isolated targetQuantityDefinition in selected accountConversionFraction of starting reactant consumedSelectivityFraction routed to the specified target pathIsolated yieldTarget recovered / theoretical targetReaction progress, target formation and recovery are distinct.
Original definition organizer; no universal selectivity formula is imposed on reactions with different stoichiometric bases.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What evidence separates poor target formation from losing target during purification?

Visual reasoning — after your attempt

Analyze reaction composition before isolation and compare it with the amount and purity recovered afterward, using compatible material balances.

A reactant can be consumed in side reactions that do not produce the desired product. Low recovered yield can also reflect incomplete reaction or loss during separation. A mass result alone does not uniquely identify which mechanism reduced target yield; compare chemical and recovery evidence.

If reactant disappears but target product is low, investigate side products and handling loss separately. Do not automatically conclude the limiting reactant calculation was wrong.

Pause and explain

What evidence distinguishes chemical loss from recovery loss?

Check your reasoning after answering

Analyze remaining reactants and side products, and compare product before and after separation using appropriate supplied measurements.

Apply the whole unit · Find a limiting amount and percent yield

Illustrative model: N₂+3H₂→2NH₃; initial N₂=1.00 mol and H₂=2.00 mol. M(NH₃)=17.0 g/mol. Pure recovered NH₃ is 18.0 g.

Your task

Identify the limiting reactant, calculate theoretical ammonia and excess nitrogen remaining, find percent yield, and propose evidence needed to explain the yield shortfall.

What your response should show

  • Choose the appropriate chemical model or relationship and state its conditions.
  • Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
  • Use supplied evidence to justify the conclusion; distinguish observation from inference.
  • Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
  1. N₂ can support 2.00 mol NH₃; H₂ can support 2.00×2/3=1.333 mol.

  2. H₂ limits; theoretical ammonia≈22.7 g.

  3. N₂ consumed=2.00/3=0.667 mol; leftover≈0.333 mol.

  4. Percent yield=18.0/(1.333…×17.0)×100≈79.4%.

Hydrogen is limiting despite the one-versus-two initial mole numbers because the reaction requires three hydrogen per nitrogen. The predicted leftovers assume complete consumption through this target reaction. A 79.4% recovered yield can result from incomplete conversion, side reactions, or separation loss; additional evidence is needed to distinguish them.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require both product-capacity calculations before naming the limiting reactant. Keep unrounded intermediate amounts and ask for a chemically plausible explanation of low yield. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.

The limiting reactant runs out firstRecipe: 1 frame + 2 wheels → 1 bicycleYou have4 frames, 6 wheelsYou can build3 bicycles1 frame left over (excess)Wheels run out first, so wheels are the limiting reactant — they cap the build at 3.Percent yield asks: of those 3 possible, how many did you actually finish? (actual ÷ theoretical × 100%)
With 4 frames but only 6 wheels, you can build just 3 bicycles — the wheels are the 'limiting reactant' and a frame is left over in excess. Percent yield then compares what you actually built to that maximum of 3.

The key idea

The limiting reactant runs out first and sets the theoretical yield (the maximum product possible); percent yield = (actual yield ÷ theoretical yield) × 100% measures how much of that maximum a real reaction actually delivers.

Percent Yield

percent yield = (actual yield ÷ theoretical yield) × 100%

actual yield is what you collect in the lab; theoretical yield is the maximum predicted by stoichiometry from the limiting reactant. Rearranged, actual yield = percent yield × theoretical yield.

Worked examples

Example 1

For N₂ + 3H₂ → 2NH₃, you start with 2.0 mol of N₂ and 3.0 mol of H₂. Which reactant is limiting, and how many moles of NH₃ can form?
  1. Let each reactant predict the product. From N₂: 2.0 mol N₂ × (2 mol NH₃ ÷ 1 mol N₂) = 4.0 mol NH₃.
  2. From H₂: 3.0 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 2.0 mol NH₃.
  3. The smaller prediction wins: H₂ makes only 2.0 mol, so H₂ is the limiting reactant and 2.0 mol of NH₃ forms (the N₂ is in excess).
Answer: H₂ is limiting; 2.0 mol of NH₃ forms.

Example 2 · Worked application

For 2H₂+O₂→2H₂O, supply 4 mol H₂ and 1 mol O₂. Find limiting reactant and maximum water.
  1. One mole O₂ needs 2 mol H₂, available in excess.
  2. O₂ therefore limits reaction.
  3. Maximum water is 1×2=2 mol; 2 mol H₂ remain in the ideal model.
Answer: O₂ limiting; 2 mol H₂O and 2 mol leftover H₂.
Check the reasoning

Compare required ratios, not just smaller starting mole count.

Example 3

A reaction has a theoretical yield of 40.0 g of product, but only 34.0 g is actually collected. What is the percent yield?
  1. Write the formula: percent yield = (actual yield ÷ theoretical yield) × 100%.
  2. Substitute the values: (34.0 g ÷ 40.0 g) × 100%.
  3. Compute: 0.850 × 100% = 85.0%.
Answer: 85.0%

Example 4 · Find and fix the mistake

Context: For 2H₂+O₂→2H₂O, supply 4 mol H₂ and 1 mol O₂. Find limiting reactant and maximum water. A student claims: “The smaller mass is always the limiting reactant.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Compare required ratios, not just smaller starting mole count.
  2. Limiting status depends on stoichiometric demand and mole amounts, not mass alone.
Answer: Limiting status depends on stoichiometric demand and mole amounts, not mass alone.
Check the reasoning

Compare required ratios, not just smaller starting mole count.

Example 5 · Reason from the visual model

Percent yield compares measured target recovery with a theoretical referencePercent yield compares measured target recovery with a theoretical referenceMass (g)Theoretical product10Recovered target8.2Percent yield = 8.2/10 ×100 = 82%Wet or impure recovered material can inflate the apparent yield.
Original selected product masses; recovered target identity and dryness are assumptions to be evaluated, not supplied assay results.
Does an 82% isolated yield prove that exactly 18% of the reactants never reacted?
  1. Percent yield is actual recovered target amount divided by the theoretical amount times 100%. The comparison must use compatible quantities and product identity. Transfer losses, incomplete conversion and competing reactions can lower recovery, while water or contamination can inflate a gross recovered mass. A value above 100% usually indicates a measurement, purity or model problem rather than creation of extra atoms. A single yield number cannot distinguish all causes; examine the product assay and procedure.
  2. No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.
Answer: No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.
Check the reasoning

Original selected product masses; recovered target identity and dryness are assumptions to be evaluated, not supplied assay results.

Avoid these mistakes

Check your understanding

N₂+3H₂→2NH₃ with 2molN₂ and 3molH₂: predict product.

Discuss the answer

H₂ limits; it supports 2molNH₃ and consumes 1molN₂, leaving 1molN₂. Compare potential products, not raw amounts.

Return to the phenomenon

Return to the initial case. Why is the reactant with less starting mass not automatically limiting? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

The smaller product capacity sets the theoretical extent. Compare balanced mole demands, then calculate leftovers and distinguish formation from recovery.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why is the reactant with less starting mass not automatically limiting? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the chemical mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep species identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

The smaller product capacity sets the theoretical extent. Compare balanced mole demands, then calculate leftovers and distinguish formation from recovery.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

For N₂+3H₂→2NH₃, begin with 1.50 mol N₂ and 3.00 mol H₂. Find limiting reactant, theoretical NH₃, remaining N₂ and yield if 1.60 mol NH₃ is isolated.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

H₂ limits. It uses 1.00 mol N₂, forms 2.00 mol NH₃ and leaves 0.50 mol N₂ in the complete model. Isolated yield is 1.60/2.00×100=80.0%; that number alone does not identify the cause of lost target.

Words to know

limiting reactant
The reactant that runs out first, capping the amount of product a reaction can make.
excess reactant
A reactant that is left over after the limiting reactant is completely used up.
theoretical yield
The maximum amount of product predicted by stoichiometry, assuming the reaction goes perfectly and the limiting reactant is fully converted.
actual yield
The amount of product actually collected when the reaction is run in the real world.
percent yield
The actual yield expressed as a percentage of the theoretical yield: (actual ÷ theoretical) × 100%.

The limiting reactant isn't just 'the one you have less of'

The most dangerous shortcut in this lesson is assuming the limiting reactant is simply whichever one you have the least of. The mole ratio gets a vote too. Suppose 2H₂ + O₂ → 2H₂O and you start with 4 moles of hydrogen but only 3 moles of oxygen. Oxygen is the smaller pile, so it looks limiting — yet the recipe needs just 1 oxygen for every 2 hydrogens, so your 4 moles of H₂ require only 2 moles of O₂. You have 3, which means oxygen is actually in excess and hydrogen runs out first. Always let each reactant calculate how much product it could make; the smallest result wins, no matter which reactant is physically the smaller amount.

Why real reactions fall short of the promise

Theoretical yield is a promise made on paper, and reality keeps only part of it — but why? One major reason is that many reactions are reversible: as products build up, some convert back into reactants, and the reaction stalls at a balance point called equilibrium before all the limiting reactant is consumed. Other culprits are side reactions, where reactants wander off to make unwanted products, and plain physical loss — product left clinging to filter paper, smeared on glassware, or discarded during purification. Each leak drags the actual yield below the theoretical maximum. Percent yield is really a report card that summarizes all of these imperfections in a single, honest number.

When 90% isn't good enough: yields multiply

A single reaction at 90% yield sounds excellent — until you learn that complex molecules like medicines are built in long chains of reactions, and the step yields multiply together. Run ten steps in a row, each a respectable 90%, and the overall yield is 0.90 raised to the tenth power, which comes to only about 35%. Nearly two-thirds of your starting material vanishes across the sequence, even though no single step ever looked wasteful. This compounding is why designing a short synthetic route matters as much as making each step efficient, and why chemists celebrate a clever reaction that shaves even a few steps off the path to a valuable drug.

Beyond percent yield: atom economy and green chemistry

Percent yield answers 'how much of the possible product did I actually get?' But a modern reaction is judged by a second question too: 'how much of my starting material ended up in the product rather than in the waste bin?' That measure is called atom economy, introduced by the chemist Barry Trost in 1991. A reaction can boast a 95% yield and still be wasteful if half of every reactant molecule is destined to become a useless byproduct. A cornerstone of green chemistry, atom economy pushes chemists toward reactions in which nearly every atom of the reactants is built into the desired product — cutting waste, cost, and pollution all at once.

Can a percent yield be more than 100%?

In theory, no — you cannot collect more product than the limiting reactant is able to make, so a percent yield above 100% is impossible in principle. In practice, students and even seasoned professionals sometimes calculate one anyway, and it is always a red flag rather than a triumph. The usual culprit is an impure product: crystals still damp with water or solvent, or laced with leftover reactant, weigh more than the pure substance would, inflating the measured actual yield. A percent yield of 105% is chemistry's way of telling you to dry and purify your product more thoroughly and then weigh it again — the extra mass is not product, it is a mistake wearing product's clothes.

In the real world

Chemical companies obsess over percent yield because every point lost is wasted raw material and money; a synthesis that jumps from 70% to 90% yield can turn an unprofitable drug into a viable one.

Exit check

N₂+3H₂→2NH₃ with 2molN₂ and 3molH₂: predict product.

Check your explanation after answering

H₂ limits; it supports 2molNH₃ and consumes 1molN₂, leaving 1molN₂. Compare potential products, not raw amounts.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

The limiting reactant in a chemical reaction is the reactant that…
  1. has the largest molar mass
  2. is added in the greatest amount
  3. runs out first and limits how much product forms
  4. is always left over at the end
Answer and explanation

runs out first and limits how much product forms

The limiting reactant is used up first, setting the ceiling on how much product can be made.

Problem 2

A reaction has a theoretical yield of 50.0 g but an actual yield of 45.0 g. What is the percent yield?
Answer and explanation

90 %

percent yield = (45.0 ÷ 50.0) × 100% = 90%.

Problem 3

For 2H₂ + O₂ → 2H₂O, you have 4 mol of H₂ and 4 mol of O₂. Which is the limiting reactant?
  1. H₂
  2. O₂
  3. both are limiting
  4. neither — they are in perfect ratio
Answer and explanation

H₂

The ratio needs 2 H₂ per 1 O₂. 4 mol H₂ requires only 2 mol O₂ (so O₂ is in excess); 4 mol O₂ would need 8 mol H₂ (not available). H₂ runs out first, so H₂ is limiting.

Problem 4

For N₂ + 3H₂ → 2NH₃, you have 2.0 mol of N₂ and 3.0 mol of H₂. How many moles of NH₃ can form?
Answer and explanation

2 mol

N₂ could make 4.0 mol NH₃, but H₂ could make only 3.0 × (2 ÷ 3) = 2.0 mol. H₂ is limiting, so 2.0 mol NH₃ forms.

Problem 5

A student predicts a theoretical yield of 25.0 g but isolates only 20.0 g of product. What is the percent yield?
Answer and explanation

80 %

percent yield = (20.0 ÷ 25.0) × 100% = 80%.

Problem 6

A reaction has a theoretical yield of 8.00 g. If it runs at 75.0% yield, what is the actual yield?
Answer and explanation

6 g

actual yield = percent yield × theoretical yield = 0.750 × 8.00 g = 6.00 g.

Problem 7

Theoretical yield is 25 g and actual isolated yield is 20 g. Find percent yield. Show the reasoning, not just the conclusion.
Hints

Compare required ratios, not just smaller starting mole count.

Answer and explanation

20/25×100%=80%.

Problem 8

A recovered product weighs more than theoretical yield. What should be checked?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Wetness, impurities, measurement, reaction model or omitted inputs; apparent yield above 100% does not prove extra atoms were created.

Problem 9

Percent yield compares measured target recovery with a theoretical referencePercent yield compares measured target recovery with a theoretical referenceMass (g)Theoretical product10Recovered target8.2Percent yield = 8.2/10 ×100 = 82%Wet or impure recovered material can inflate the apparent yield.
Original selected product masses; recovered target identity and dryness are assumptions to be evaluated, not supplied assay results.
Interpret the supplied visual. Does an 82% isolated yield prove that exactly 18% of the reactants never reacted?
Answer and explanation

No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.

Percent yield is actual recovered target amount divided by the theoretical amount times 100%. The comparison must use compatible quantities and product identity. Transfer losses, incomplete conversion and competing reactions can lower recovery, while water or contamination can inflate a gross recovered mass. A value above 100% usually indicates a measurement, purity or model problem rather than creation of extra atoms. A single yield number cannot distinguish all causes; examine the product assay and procedure.

No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

For N₂+3H₂→2NH₃, begin with 1.50 mol N₂ and 3.00 mol H₂. Find limiting reactant, theoretical NH₃, remaining N₂ and yield if 1.60 mol NH₃ is isolated.
Answer and explanation

H₂ limits. It uses 1.00 mol N₂, forms 2.00 mol NH₃ and leaves 0.50 mol N₂ in the complete model. Isolated yield is 1.60/2.00×100=80.0%; that number alone does not identify the cause of lost target.

H₂ limits. It uses 1.00 mol N₂, forms 2.00 mol NH₃ and leaves 0.50 mol N₂ in the complete model. Isolated yield is 1.60/2.00×100=80.0%; that number alone does not identify the cause of lost target.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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