Stoichiometry & Yield — Chapter Practice, Grade 11

Calculate reacting amounts, identify shortages, and interpret production results.

Problem 1

In the balanced equation N₂ + 3H₂ → 2NH₃, what is the mole ratio of H₂ to NH₃?
  1. 3 : 1
  2. 3 : 2
  3. 1 : 1
  4. 2 : 3
Answer and explanation

3 : 2

The coefficients give H₂ : NH₃ = 3 : 2, read straight from the balanced equation.

Problem 2

For 2H₂ + O₂ → 2H₂O, how many moles of water form from 4.0 mol of hydrogen reacting with excess oxygen?
Answer and explanation

4 mol

The mole ratio of H₂O to H₂ is 2 : 2 = 1 : 1, so 4.0 mol H₂ → 4.0 mol H₂O.

Problem 3

For N₂ + 3H₂ → 2NH₃, how many moles of ammonia form from 9.0 mol of hydrogen (with excess nitrogen)?
Answer and explanation

6 mol

9.0 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 6.0 mol NH₃.

Problem 4

For C + O₂ → CO₂, what mass of carbon dioxide is produced when 1.00 mol of carbon burns completely? (Molar mass CO₂ = 44.01 g/mol.)
Answer and explanation

44.01 g

The mole ratio of CO₂ to C is 1 : 1, so 1.00 mol C → 1.00 mol CO₂, and 1.00 mol × 44.01 g/mol = 44.01 g.

Problem 5

For 2H₂ + O₂ → 2H₂O, what mass of water forms from 4.04 g of hydrogen reacting with excess oxygen? (Molar masses: H₂ = 2.02 g/mol, H₂O = 18.02 g/mol.)
Answer and explanation

36.04 g

4.04 g ÷ 2.02 g/mol = 2.00 mol H₂; ratio H₂O:H₂ = 1:1 gives 2.00 mol H₂O; 2.00 mol × 18.02 g/mol = 36.04 g.

Problem 6

For N₂+3H₂→2NH₃, find NH₃ from 1.5 mol N₂ with excess H₂. Show the reasoning, not just the conclusion.
Hints

Mole ratios come from coefficients, not molar masses.

Answer and explanation

1.5×2/1=3.0 mol NH₃ theoretical yield.

Problem 7

Why must a grams-to-grams problem usually pass through moles?
Hints

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

Answer and explanation

Balanced coefficients relate particle counts/moles, so masses must be converted using each substance's molar mass.

Problem 8

Theoretical product is conditional on purity, partners and reaction extentTheoretical product is conditional on purity, partners and reaction extentConditions behind a theoretical amountCorrect balanced reaction and molar massesStarting purity and available reacting amountSufficient partner / stated completeness / no competing pathA calculation predicts under assumptions; it does not measure yield.
Original assumption organizer, not a complete kinetic/equilibrium calculation or a general promise of quantitative conversion.
Interpret the supplied visual. What changes if only 80% of a 10.0 g starting sample is the specified reactant?
Answer and explanation

Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.

A theoretical yield is the product amount predicted by a stated chemical model and available reacting quantities. An impure sample contains less target reactant than its total mass, an insufficient partner can limit conversion, and an equilibrium or competing pathway can reduce the modeled target formation. These are distinct limitations, not reasons to abandon conservation. State the reaction, purity, available amounts and completeness assumptions before presenting a number as an experimental outcome.

Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.

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 9

For 2Mg+O₂→2MgO, use Mg=24.3 and MgO=40.3 g/mol. Find theoretical MgO from 4.86 g pure Mg with excess oxygen. Explain the mass increase.
Answer and explanation

4.86/24.3=0.200 mol Mg. The 1:1 ratio gives 0.200 mol MgO and 0.200×40.3=8.06 g MgO. The extra mass comes from incorporated oxygen; the complete mass account includes it.

4.86/24.3=0.200 mol Mg. The 1:1 ratio gives 0.200 mol MgO and 0.200×40.3=8.06 g MgO. The extra mass comes from incorporated oxygen; the complete mass account includes it.

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.

Problem 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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.

Problem 20

Find moles in 0.250 L of 0.400 M solution.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

0.1 mol

.400×.250=.100.

Problem 21

Why can equal solution volumes require unequal reacting amounts?
Answer and explanation

Their molarities and reaction coefficients can differ.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 22

Find mL of .20 M NaOH needed for .0040 mol HCl in a 1:1 reaction.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

20 mL

.0040/.20=.020L=20 mL.

Problem 23

Compare dilution and neutralization equations.
Answer and explanation

Dilution preserves solute moles; neutralization transforms species according to a balanced equation.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 24

Find solute moles in 0.025 L of 0.40 M solution. Show the reasoning, not just the conclusion.
Hints

Use liters in n=MV.

Answer and explanation

n=0.40×0.025=0.010 mol.

Problem 25

Equal volumes of 0.10 M Ag⁺ and 0.20 M Cl⁻ react 1:1. Which ion limits?
Hints

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

Answer and explanation

Ag⁺, because equal volumes give half as many moles of Ag⁺ as Cl⁻.

Problem 26

Reaction bookkeeping comes before calculating the final concentrationReaction bookkeeping comes before calculating the final concentrationMixing accountBefore reactionAfter reaction/mixingSolute amountComputed in each portionSubtract what reactsSolution volumeTwo measured portionsUse measured final VConcentrationEach initial cRemaining n / final VAssuming additive volumes is a stated approximation.
Original bookkeeping organizer; it supplies the calculation order rather than numerical solution-volume measurements.
Interpret the supplied visual. Why is dividing initial acid moles by total mixed volume not enough after adding a reacting base?
Answer and explanation

Some acid is consumed. Reaction bookkeeping must determine the remaining amount before the final-volume division.

When solutions mix, a reacting ion can be consumed while spectator ions remain. Calculate each initial amount, apply stoichiometry, then divide remaining amount by the final solution volume. Adding volumes is a common dilute-solution school approximation, not an exact law for every mixture. The concentration of a spectator can decrease by dilution without the spectator being consumed. A final mixture can contain several species with different concentrations, so keep a separate inventory for each named species.

Some acid is consumed. Reaction bookkeeping must determine the remaining amount before the final-volume division.

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 27

Mix 10.0 mL of 0.200 M AgNO₃ with 15.0 mL of 0.100 M NaCl. Assume complete AgCl precipitation and additive volumes. Find limiting ion, AgCl moles and remaining Ag⁺ concentration.
Answer and explanation

Ag⁺ starts at 0.00200 mol and Cl⁻ at 0.00150 mol, so chloride limits. AgCl forms 0.00150 mol; remaining Ag⁺ is 0.00050 mol in 0.0250 L, or 0.0200 M under the stated model.

Ag⁺ starts at 0.00200 mol and Cl⁻ at 0.00150 mol, so chloride limits. AgCl forms 0.00150 mol; remaining Ag⁺ is 0.00050 mol in 0.0250 L, or 0.0200 M under the stated model.

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.

Browse all courses