Reactions, Moles & Composition — Chapter Practice, Grade 11

Account for atoms, charge, particle counts, and measured composition.

Problem 1

Two or more simpler substances combine to form a single product, A + B → AB. What type of reaction is this?
  1. synthesis
  2. decomposition
  3. single replacement
  4. combustion
Answer and explanation

synthesis

Several substances combining into one product is the defining pattern of a synthesis (combination) reaction.

Problem 2

Which equation represents a combustion reaction?
  1. CaCO₃ → CaO + CO₂
  2. CH₄ + 2O₂ → CO₂ + 2H₂O
  3. AgNO₃ + NaCl → AgCl + NaNO₃
  4. 2Na + Cl₂ → 2NaCl
Answer and explanation

CH₄ + 2O₂ → CO₂ + 2H₂O

A fuel (methane) reacting with oxygen to give carbon dioxide and water is complete combustion.

Problem 3

Classify the reaction 2KClO₃ → 2KCl + 3O₂.
  1. synthesis
  2. decomposition
  3. single replacement
  4. double replacement
Answer and explanation

decomposition

A single compound breaking apart into simpler substances is decomposition (AB → A + B).

Problem 4

In the single-replacement reaction Zn + 2HCl → ZnCl₂ + H₂, how many molecules of hydrogen gas (H₂) are produced for every one atom of zinc that reacts?
Answer and explanation

1 molecules

The coefficient in front of H₂ is 1, so one zinc atom releases one H₂ molecule.

Problem 5

Solid magnesium is burned in oxygen gas. Predict the product, write the type of reaction, and give the balanced equation.
Answer and explanation

It is a synthesis reaction. Magnesium combines with oxygen to form magnesium oxide, MgO. The balanced equation is 2Mg + O₂ → 2MgO.

Problem 6

Which of these is a single-replacement reaction, in which one element displaces another from a compound?
  1. CaCO₃ → CaO + CO₂
  2. 2H₂ + O₂ → 2H₂O
  3. Fe + CuSO₄ → FeSO₄ + Cu
  4. HCl + NaOH → NaCl + H₂O
Answer and explanation

Fe + CuSO₄ → FeSO₄ + Cu

Iron (a lone element) replaces copper in copper sulfate, matching A + BC → AC + B. Because iron is more reactive than copper, the displacement actually occurs.

Problem 7

A hydrocarbon reacts with oxygen to form CO₂ and water. What broad category fits? Show the reasoning, not just the conclusion.
Hints

Use supplied products and states to justify the category.

Answer and explanation

Complete combustion under the given products; it is also a redox process.

Problem 8

What observation would support acid–base neutralization rather than prove it solely from temperature rise?
Hints

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

Answer and explanation

Reactant identities and proton-transfer/stoichiometric evidence; warming alone occurs in other processes too.

Problem 9

Aqueous outcomes depend on dissolved species and product conditionsAqueous outcomes depend on dissolved species and product conditionsAg⁺(aq) + Cl⁻(aq) → AgCl(s)Before mixingSelected solid phaseUse solubility evidence and the actual dissolved concentrations.Ag⁺Cl⁻+−+−+−
Original qualitative silver-chloride precipitation model. Charges in the solid slice are schematic; no concentration or solubility-product calculation is supplied.
Interpret the supplied visual. What does the symbol (s) add to the AgCl product account?
Answer and explanation

It identifies a separate solid phase, distinguishing precipitation from ions simply remaining in the aqueous phase.

The state symbol (aq) means a species is dissolved in water, not that a solid formula unit is moving unchanged through solution. Precipitation forms a separate solid phase when conditions favor it. A supplied solubility rule can support an introductory prediction, while concentrations and equilibrium determine the more detailed outcome. Gas formation and acid–base reactions can also change aqueous species. A visible change suggests something to investigate but does not by itself establish the exact product identity or amount.

It identifies a separate solid phase, distinguishing precipitation from ions simply remaining in the aqueous phase.

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

A student writes an exchange equation for two clear ionic solutions and predicts a solid solely because the equation is balanced. Explain the missing evidence and give two quantities that still must balance.
Answer and explanation

Balance atoms of each element and total charge. A precipitate prediction also needs appropriate solubility/concentration evidence under the conditions; balanced pattern matching alone does not establish a separate solid phase.

Balance atoms of each element and total charge. A precipitate prediction also needs appropriate solubility/concentration evidence under the conditions; balanced pattern matching alone does not establish a separate solid phase.

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 11

OpenSciEd diagram linking a solid bath bomb and water before mixing to gas bubbles and liquid after mixing, with empty magnification circles for the learner’s particle model.
This OpenSciEd template is an initial-model prompt with intentionally empty particle views; it is not a completed molecular reaction diagram. Bath-bomb matter model at visible and particle scales · OpenSciEd · 7.1 Chemical Reactions & Matter, Lesson 1, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
The source template has separate visible and magnified views before and after gas bubbles appear. State two observations it invites, two details a completed particle model needs, and why bubbles alone do not prove boiling or gas identity.
Answer and explanation

Record the visible solid/water and later bubbles or liquid. A completed model must define its species symbols and conserve atoms/material across the boundary, including dissolved and gas species. Bubbles can arise through reaction, vaporization or released dissolved gas; temperature and product evidence are needed to distinguish them.

Identify the source’s visible labels, scales and unfinished or omitted information.

Use the local complete explanations to connect the observation with an explicit mechanism.

State assumptions and a limit before extending the model.

Response criteria

  • Interpret the source symbols and boundary correctly.
  • Explain the relevant mechanism using quantities/species consistently.
  • State a meaningful omission, assumption or evidence limitation.

Problem 12

When balancing an equation, why may you change coefficients but never subscripts?
  1. Subscripts must always be 1
  2. There is no real difference between them
  3. Coefficients are easier to write
  4. Changing a subscript would change the substance into a different chemical, while a coefficient only changes how many units you have
Answer and explanation

Changing a subscript would change the substance into a different chemical, while a coefficient only changes how many units you have

A subscript defines a substance's identity (H₂O vs H₂O₂), so only coefficients — the 'how many' — can be adjusted.

Problem 13

In the balanced equation N₂ + 3H₂ → 2NH₃, how many hydrogen atoms are on each side?
Answer and explanation

6 atoms

Left: 3H₂ = 3 × 2 = 6 H. Right: 2NH₃ = 2 × 3 = 6 H. They match.

Problem 14

Which set of coefficients correctly balances ___Al + ___O₂ → ___Al₂O₃?
  1. 1, 1, 1
  2. 2, 3, 1
  3. 4, 3, 2
  4. 2, 1, 2
Answer and explanation

4, 3, 2

4Al + 3O₂ → 2Al₂O₃ gives 4 Al and 6 O atoms on both sides — the smallest whole-number balance.

Problem 15

Balance the decomposition ___KClO₃ → ___KCl + ___O₂. What coefficient belongs in front of O₂?
Answer and explanation

3

2KClO₃ → 2KCl + 3O₂ balances: K 2 = 2, Cl 2 = 2, O 6 = 6. The coefficient on O₂ is 3.

Problem 16

Explain how balancing a chemical equation is a direct application of the law of conservation of mass.
Answer and explanation

Conservation of mass says atoms are only rearranged, never created or destroyed, so the same number of each atom must appear on both sides. Balancing forces the atom counts to match, which guarantees the total mass of the reactants equals the total mass of the products.

Problem 17

Gasoline burns by the equation 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. How many molecules of O₂ are required to burn 2 molecules of octane (C₈H₁₈)?
Answer and explanation

25 molecules

The balanced coefficient on O₂ is 25: two octane molecules need 25 oxygen molecules for complete combustion.

Problem 18

Balance Al + O₂ → Al₂O₃. Show the reasoning, not just the conclusion.
Hints

Balanced coefficients are ratios, not proof of complete reaction under every condition.

Answer and explanation

4Al + 3O₂ → 2Al₂O₃; Al=4 and O=6 on both sides.

Problem 19

Why can a balanced reaction have unequal totals of molecules on its two sides?
Hints

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

Answer and explanation

Atoms are conserved, but grouping those atoms into different molecules changes molecule counts.

Problem 20

Smallest whole-number coefficients are a reporting conventionSmallest whole-number coefficients are a reporting convention4H₂ + 2O₂ → 4H₂O2H₂ + O₂ → 2H₂ODivide every coefficient by their common factor of 2.The physical ratio is unchanged.
Original coefficient-scaling comparison, not a claim that a reaction’s mechanism requires exactly this many simultaneous molecular collisions.
Interpret the supplied visual. Why can 4:2:4 and 2:1:2 describe the same reaction ratio while an associated heat value must be scaled?
Answer and explanation

The ratios are equivalent, but the thermochemical amount basis changes. Heat is extensive and must change with the equation amount.

Multiplying all coefficients by the same positive factor preserves the stoichiometric ratio and conservation. For an ordinary school balancing answer, report the smallest whole-number coefficients. A thermochemical equation may deliberately use a different scale, including a fraction, to define an energy change per specified amount. The equation’s normalization matters when relating it to moles or reaction enthalpy; reducing one coefficient alone would break the composition account. State the convention and scale the full equation consistently.

The ratios are equivalent, but the thermochemical amount basis changes. Heat is extensive and must change with the equation amount.

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 21

Balance C₃H₈ + O₂ → CO₂ + H₂O using the smallest whole-number coefficients, then show the atom counts.
Answer and explanation

C₃H₈ +5O₂ →3CO₂ +4H₂O. Counts are C:3, H:8, O:10 on each side. Formula subscripts remain unchanged.

C₃H₈ +5O₂ →3CO₂ +4H₂O. Counts are C:3, H:8, O:10 on each side. Formula subscripts remain unchanged.

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 22

Calculate the molar mass of carbon dioxide, CO₂. (C = 12.01, O = 16.00 g/mol.)
Answer and explanation

44.01 g/mol

Molar mass = 12.01 + 2(16.00) = 12.01 + 32.00 = 44.01 g/mol.

Problem 23

Avogadro's number — the number of representative particles in one mole — is approximately…
  1. 6.022 × 10²³
  2. 3.14 × 10⁰
  3. 1.602 × 10⁻¹⁹
  4. 9.81 × 10⁰
Answer and explanation

6.022 × 10²³

One mole is defined as 6.022 × 10²³ particles.

Problem 24

How many moles are in 88.02 g of carbon dioxide? (Molar mass CO₂ = 44.01 g/mol.)
Answer and explanation

2 mol

moles = mass ÷ molar mass = 88.02 g ÷ 44.01 g/mol = 2.00 mol.

Problem 25

What is the mass of 3.00 mol of water? (Molar mass H₂O = 18.02 g/mol.)
Answer and explanation

54.06 g

mass = moles × molar mass = 3.00 mol × 18.02 g/mol = 54.06 g.

Problem 26

How many molecules are in 2.00 mol of carbon dioxide? (Avogadro's number = 6.022 × 10²³ per mole.)
Answer and explanation

1.2044e+24 molecules

molecules = moles × Avogadro's number = 2.00 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.

Problem 27

Calculate the molar mass of glucose, C₆H₁₂O₆. (C = 12.01, H = 1.01, O = 16.00 g/mol.)
Answer and explanation

180.18 g/mol

6(12.01) + 12(1.01) + 6(16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol.

Problem 28

Find moles in 11.0 g CO₂ using 44.0 g/mol. Show the reasoning, not just the conclusion.
Hints

Identify the requested entity before applying the mole conversion.

Answer and explanation

11.0/44.0=0.250 mol.

Problem 29

How many oxygen atoms are in 0.50 mol O₂ using Nₐ=6.02×10²³?
Hints

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

Answer and explanation

0.50 mol molecules×2=1.0 mol O atoms, or 6.02×10²³ atoms.

Problem 30

Units connect measured mass to amount and entity countUnits connect measured mass to amount and entity count36.0 g H₂O2.00 mol H₂O÷ 18.0 g/mol2.00 mol × 6.022×10²³ molecules/mol= 1.20×10²⁴ H₂O molecules
Original rounded conversion using a supplied water molar mass; entity count precision reflects the measured mass and chosen school constants.
Interpret the supplied visual. What would be wrong with multiplying 36.0 g directly by a number in molecules/mol?
Answer and explanation

The grams would not cancel the mole denominator. First convert mass to moles with the substance’s molar mass.

Convert mass to amount using n=m/Mₘ, then multiply amount by the Avogadro constant to obtain specified-entity count. Each conversion factor must cancel the previous unit and retain the entity name. The selected 18.0 g/mol water value is a rounded school input; using it consistently gives 2.00 mol from 36.0 g and approximately 1.20×10²⁴ water molecules. Mole amount is not itself a mass or volume, so do not skip the chemical composition step.

The grams would not cancel the mole denominator. First convert mass to moles with the substance’s molar mass.

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 31

Using Mₘ(CO₂)=44.0 g/mol and Nₐ≈6.022×10²³ mol⁻¹, find moles, molecules and moles of oxygen atoms in 22.0 g CO₂.
Answer and explanation

n=22.0/44.0=0.500 mol CO₂. Molecules≈3.01×10²³. Each has two O atoms, giving 1.00 mol O atoms. Keep the entity labels through every conversion.

n=22.0/44.0=0.500 mol CO₂. Molecules≈3.01×10²³. Each has two O atoms, giving 1.00 mol O atoms. Keep the entity labels through every conversion.

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 32

With C=12 and O=16, find percent carbon in CO₂.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

27.27 %

12/44 ×100 = 27.27%.

Problem 33

Why not use mass ratios directly as formula subscripts?
Answer and explanation

Subscripts count atoms; convert mass to moles because elements have different atomic masses.

Response criteria

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

Problem 34

An empirical formula mass is 15 g/mol and the molecular molar mass is 60 g/mol. Find the subscript multiplier.
Hints

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

Answer and explanation

4

60/15=4.

Problem 35

Empirical mass is 46 g/mol and molecular mass 92. Find multiplier.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

2

92/46 = 2.

Problem 36

An empirical formula is CH₂O, mass 30 g/mol; measured molar mass is 90 g/mol. Find molecular formula. Show the reasoning, not just the conclusion.
Hints

Percentages describe mass, so convert to moles before atom ratios.

Answer and explanation

90/30=3; multiply subscripts to get C₃H₆O₃.

Problem 37

A compound has molar masses 24 g of C and 4 g of H in the analyzed portion. Use C=12 and H=1 to find empirical formula.
Hints

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

Answer and explanation

Moles C=2, H=4; ratio 1:2, giving CH₂.

Problem 38

Near-simple fractions should be tested against the evidenceNear-simple fractions should be tested against the evidenceInitial ratio: 1.00 : 1.50Multiply every term by 2Whole-number ratio: 2 : 3Do not round 1.50 to 2 before checking a simple multiplier.
Original fractional-ratio demonstration without an assigned chemical species or measurement uncertainty.
Interpret the supplied visual. Why is multiplying the entire 1:1.5 ratio by two preferable to rounding only 1.5?
Answer and explanation

It preserves the measured proportional relationship, yielding 2:3; rounding one term changes the ratio.

An empirical ratio near 1:1.5 can support a 2:3 atom ratio after multiplying all terms by two. Premature rounding to 1:2 changes the composition substantially. Look for simple fractional relationships and compare the reconstructed percent composition with the original data and uncertainty. A complicated multiplier chosen only to force an answer can overfit noisy measurements. If several formulas fit within error, report that ambiguity and request more evidence rather than treating the nearest decimal as an exact atomic count.

It preserves the measured proportional relationship, yielding 2:3; rounding one term changes the ratio.

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 39

A selected oxide sample has 70.0 g Fe and 30.0 g O. Use Fe=56.0 and O=16.0 g/mol to find the simplest formula. State one assumption.
Answer and explanation

Amounts are 1.25 mol Fe and 1.875 mol O, ratio 1:1.5, hence Fe₂O₃. Assume the analyzed sample is one oxide accounted for by those elements and the input precision supports that ratio.

Amounts are 1.25 mol Fe and 1.875 mol O, ratio 1:1.5, hence Fe₂O₃. Assume the analyzed sample is one oxide accounted for by those elements and the input precision supports that ratio.

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 40

Identify spectators in silver nitrate plus sodium chloride.
Answer and explanation

Na⁺ and NO₃⁻ remain dissolved unchanged.

Response criteria

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

Problem 41

Why leave a solid intact in a net ionic equation?
Answer and explanation

It is not represented as independently mobile aqueous ions.

Response criteria

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

Problem 42

In Ag⁺+Cl⁻→AgCl(s), how many mol Ag⁺ react with 0.30 mol Cl⁻ if silver ions are sufficient?
Hints

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

Answer and explanation

0.3 mol

The ratio is 1:1.

Problem 43

Why is swapping ions not sufficient evidence of reaction?
Answer and explanation

If all species remain unchanged dissolved ions, no net chemical transformation is represented.

Response criteria

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

Problem 44

Given BaSO₄ insoluble, write the net ionic precipitation equation from Ba²⁺ and SO₄²⁻. Show the reasoning, not just the conclusion.
Hints

Atoms and total charge must balance.

Answer and explanation

Ba²⁺(aq)+SO₄²⁻(aq)→BaSO₄(s); total charge is zero on each side.

Problem 45

Why is a solid precipitate not split into free aqueous ions in the net equation?
Hints

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

Answer and explanation

Its state is a solid lattice; representing free aqueous ions would describe a different state.

Problem 46

Net ionic equations retain both composition and charge conservationNet ionic equations retain both composition and charge conservationNet processAtom checkCharge checkAg⁺ + Cl⁻ → AgCl(s)Ag 1; Cl 1 eachside0 → 0H₃O⁺ + OH⁻ → 2H₂OH 4; O 2 each side0 → 0H⁺(aq) is often shorthand for hydrated proton species.Keep the chosen water/proton notation consistent.
Original selected conservation table; hydronium notation is a simplified aqueous proton account, not a complete hydration-cluster model.
Interpret the supplied visual. Why does H₃O⁺+OH⁻ produce two water molecules rather than one in this explicit notation?
Answer and explanation

There are four hydrogen and two oxygen atoms on the reactant side; two H₂O preserve them along with total charge.

After cancellation, check atoms and total charge again. A precipitation equation can be neutral overall while containing charged reactants. For acid–base accounts, H⁺(aq) is a common shorthand; an explicit hydronium model uses H₃O⁺ and includes the corresponding water molecules. Do not mix notations in a way that loses hydrogen or oxygen atoms. The net equation identifies the dominant modeled transformation, not every short-lived microscopic intermediate or all equilibria in the solution.

There are four hydrogen and two oxygen atoms on the reactant side; two H₂O preserve them along with total charge.

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 47

Write the net ionic equation when suitable aqueous BaCl₂ and Na₂SO₄ produce BaSO₄(s). Name spectators and verify charge. This is a supplied-species exercise, not a home experiment.
Answer and explanation

Ba²⁺(aq)+SO₄²⁻(aq)→BaSO₄(s). Na⁺ and Cl⁻ are spectators in the selected account. One Ba and one sulfate group are conserved and total charge is zero on each side.

Ba²⁺(aq)+SO₄²⁻(aq)→BaSO₄(s). Na⁺ and Cl⁻ are spectators in the selected account. One Ba and one sulfate group are conserved and total charge is zero on each side.

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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