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Quiz: Chemical Reactions and Equations

Test your understanding of chemical equations, reaction types, stoichiometry, and redox reactions with these questions.

1. A student writes the following equation for the combustion of butane: \(\ce{C4H10 + O2 -> CO2 + H2O}\). What is the correct set of coefficients that balances this equation?

  1. 2, 13, 8, 10
  2. 1, 5, 4, 5
  3. 1, 6, 4, 4
  4. 2, 10, 8, 10
Show Answer

The correct answer is A. Balancing \(\ce{C4H10}\) combustion: balance C first (gives 4 \(\ce{CO2}\)), then H (gives 5 \(\ce{H2O}\)), then O: right side has \(4 \times 2 + 5 \times 1 = 13\) oxygen atoms, requiring \(\frac{13}{2}\) \(\ce{O2}\). Multiplying through by 2 gives: \(\ce{2C4H10 + 13O2 -> 8CO2 + 10H2O}\). Verification: C: 8 = 8, H: 20 = 20, O: 26 = 26. Option B attempts single-molecule balancing that fails to account for oxygen correctly.

Concept Tested: Balancing Equations / Combustion Reactions

2. Which of the following best classifies the reaction: \(\ce{Fe(s) + CuSO4(aq) -> FeSO4(aq) + Cu(s)}\)?

  1. Synthesis reaction, because two substances combine to form a product
  2. Double replacement reaction, because two ionic compounds exchange ions
  3. Single replacement reaction, because iron displaces copper from solution
  4. Decomposition reaction, because copper sulfate breaks apart
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The correct answer is C. This is a single replacement reaction: an uncombined element (Fe) displaces another element (Cu) from a compound (\(\ce{CuSO4}\)). The general pattern A + BC → AC + B is clearly present. Iron is above copper in the activity series, so this reaction proceeds spontaneously. A double replacement would require two ionic compounds exchanging partners; a synthesis would produce one product from multiple reactants; a decomposition would break one compound into simpler substances.

Concept Tested: Single Replacement Reactions / Activity Series

3. When aqueous solutions of \(\ce{AgNO3}\) and \(\ce{NaCl}\) are mixed, a white precipitate forms. Which net ionic equation correctly represents this reaction?

  1. \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)
  2. \(\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO3-(aq)}\)
  3. \(\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}\)
  4. \(\ce{Na+(aq) + NO3-(aq) -> NaNO3(aq)}\)
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The correct answer is C. The net ionic equation is obtained by writing the complete ionic equation, identifying spectator ions (\(\ce{Na+}\) and \(\ce{NO3-}\), which appear unchanged on both sides), and canceling them. What remains — \(\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}\) — shows only the particles that actually participate in the chemical change. Option A is the molecular equation (not net ionic). Option B is the complete ionic equation (spectator ions not yet removed). Option D shows only the spectator ions, which do not react.

Concept Tested: Net Ionic Equations / Spectator Ions / Precipitation Reactions

4. A student reacts 56.0 g of iron (\(\ce{Fe}\), molar mass = 55.85 g/mol) with excess sulfur to form iron(II) sulfide (\(\ce{FeS}\), molar mass = 87.91 g/mol). What is the theoretical yield of \(\ce{FeS}\)?

  1. 55.9 g
  2. 126 g
  3. 44.0 g
  4. 87.9 g
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The correct answer is D. The balanced equation is \(\ce{Fe(s) + S(s) -> FeS(s)}\) — a 1:1 mole ratio. Moles of \(\ce{Fe}\) = \(\frac{56.0 \text{ g}}{55.85 \text{ g/mol}} = 1.002 \text{ mol}\). Since sulfur is in excess, Fe is the limiting reagent. Moles of \(\ce{FeS}\) produced = 1.002 mol. Theoretical yield = \(1.002 \text{ mol} \times 87.91 \text{ g/mol} \approx 88.1 \text{ g} \approx 87.9 \text{ g}\). Option A incorrectly uses the molar mass of Fe as the product mass, and option B adds the molar masses of the two reactants together as a mistaken shortcut.

Concept Tested: Stoichiometry / Mass-to-Mass Calculations / Theoretical Yield

5. In an experiment, 40.0 g of \(\ce{N2}\) and 15.0 g of \(\ce{H2}\) are combined according to \(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\). Which reactant is the limiting reagent, and what is the theoretical yield of \(\ce{NH3}\)?

  1. \(\ce{H2}\) is limiting; theoretical yield = 84.6 g \(\ce{NH3}\)
  2. \(\ce{N2}\) is limiting; theoretical yield = 48.7 g \(\ce{NH3}\)
  3. Both reactants are consumed equally; theoretical yield = 136 g \(\ce{NH3}\)
  4. \(\ce{N2}\) is limiting; theoretical yield = 97.2 g \(\ce{NH3}\)
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The correct answer is B. To find the limiting reagent, divide each reactant's moles by its stoichiometric coefficient. Moles of \(\ce{N2}\) = \(40.0/28.02 = 1.428\) mol; divided by coefficient 1 → 1.428. Moles of \(\ce{H2}\) = \(15.0/2.016 = 7.44\) mol; divided by coefficient 3 → 2.48. \(\ce{N2}\) has the smaller quotient (1.428 < 2.48), so \(\ce{N2}\) is the limiting reagent. Theoretical yield = \(1.428 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} \times 17.03 \text{ g/mol} = 48.7 \text{ g NH}_3\). Option A shows the result if \(\ce{H2}\) were incorrectly identified as the limiting reagent.

Concept Tested: Limiting Reagent / Theoretical Yield

6. A student performs a reaction and collects 31.5 g of product. The theoretical yield is 42.0 g. What is the percent yield?

  1. 75.0%
  2. 33.3%
  3. 57.5%
  4. 133%
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The correct answer is A. Percent yield = \(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{31.5 \text{ g}}{42.0 \text{ g}} \times 100 = 75.0\%\). Option B results from inverting the fraction (42.0/31.5 × 100 = 133%, which is option D — a percent yield above 100% signals experimental error such as incomplete drying or contamination). Option C is a plausible but incorrect distractor. A 75% yield is typical for a well-performed but not optimized laboratory reaction.

Concept Tested: Percent Yield / Actual Yield

7. What is the oxidation state of chromium (Cr) in the dichromate ion \(\ce{Cr2O7^{2-}}\)?

  1. +3
  2. +4
  3. +7
  4. +6
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The correct answer is D. Using oxidation state rules: oxygen is −2, and there are 7 oxygens, contributing −14. The overall charge of the ion is −2. Let each Cr have oxidation state \(x\): \(2x + (−14) = −2\), so \(2x = 12\), therefore \(x = +6\). Chromium is in the +6 state in dichromate, the same state as in chromate (\(\ce{CrO4^{2-}}\)). Option C (+7) is the oxidation state of Mn in permanganate (\(\ce{MnO4-}\)), a very common confusion. Option A (+3) is the most stable and common oxidation state of Cr in aqueous solution, but not in this oxyanion.

Concept Tested: Assigning Oxidation States / Redox Reactions

8. In the redox reaction \(\ce{2Mg(s) + O2(g) -> 2MgO(s)}\), which statement correctly identifies the oxidizing agent and the change it undergoes?

  1. Magnesium is the oxidizing agent; it loses electrons and its oxidation state increases from 0 to +2.
  2. Oxygen is the oxidizing agent; it gains electrons and its oxidation state decreases from 0 to −2.
  3. Magnesium is the oxidizing agent; it gains electrons and its oxidation state decreases from +2 to 0.
  4. Oxygen is the oxidizing agent; it loses electrons and its oxidation state increases from −2 to 0.
Show Answer

The correct answer is B. The oxidizing agent accepts electrons and is itself reduced. Oxygen changes from oxidation state 0 (in \(\ce{O2}\)) to −2 (in \(\ce{MgO}\)) — a decrease in oxidation state, consistent with gaining electrons (reduction). Magnesium changes from 0 to +2 — an increase, meaning it loses electrons (oxidation), making Mg the reducing agent. The memory device OIL RIG (Oxidation Is Loss, Reduction Is Gain) confirms: O gains electrons, so O is reduced; the substance reduced is the oxidizing agent.

Concept Tested: Oxidizing Agents / Reducing Agents / Redox Reactions

9. A student titrates 25.0 mL of \(\ce{HCl(aq)}\) of unknown concentration with 0.150 M \(\ce{NaOH}\). At the equivalence point, 32.0 mL of \(\ce{NaOH}\) has been added. What is the concentration of the \(\ce{HCl}\) solution?

  1. 0.117 M
  2. 0.150 M
  3. 0.192 M
  4. 0.480 M
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The correct answer is C. For a 1:1 strong acid–strong base reaction, \(M_a V_a = M_b V_b\). Solving: \(M_{HCl} = \frac{M_{NaOH} \times V_{NaOH}}{V_{HCl}} = \frac{(0.150 \text{ M})(32.0 \text{ mL})}{25.0 \text{ mL}} = 0.192 \text{ M}\). Option B simply uses the titrant concentration without accounting for the volume ratio. Option D results from inverting the volume ratio (dividing 25.0 by 32.0 rather than 32.0 by 25.0). Titration calculations always require matching the volume of titrant used against the volume of analyte taken.

Concept Tested: Titration / Solution Stoichiometry / Equivalence Point

10. When balancing the redox reaction \(\ce{MnO4-(aq) + Fe^{2+}(aq) -> Mn^{2+}(aq) + Fe^{3+}(aq)}\) in acidic solution using the half-reaction method, what coefficient appears in front of \(\ce{Fe^{2+}}\) in the final balanced equation?

  1. 1
  2. 3
  3. 5
  4. 8
Show Answer

The correct answer is C. The reduction half-reaction for \(\ce{MnO4-}\) in acidic solution is: \(\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}\) (5 electrons gained). The oxidation half-reaction for \(\ce{Fe^{2+}}\) is: \(\ce{Fe^{2+} -> Fe^{3+} + e-}\) (1 electron lost). To equalize electrons transferred, multiply the iron half-reaction by 5. This gives \(\ce{5Fe^{2+} -> 5Fe^{3+} + 5e-}\), so the coefficient in front of \(\ce{Fe^{2+}}\) in the final balanced equation is 5. Option D (8) is the coefficient for \(\ce{H+}\) in the same reaction, a common mix-up.

Concept Tested: Balancing Redox Equations / Half-Reaction Method