Exercise Questions

Question 1

Unknown Compound Combustion: A 6.0 g compound produces 8.8 g of CO2 and 3.6 g of H2O upon combustion. What is its empirical formula?

  • (A) CH
  • (B) CH2O
  • (C) C2H4O
  • (D) CH4O
View Answer
Explanation: The combustion products indicate the empirical formula is CH2O.
Correct Answer: (B)

Question 2

Ammonia Formation: N2 + 3H2 → 2NH3. Calculate the number of molecules of ammonia formed when 44.8 L of nitrogen reacts with 6.0 g of hydrogen at STP.

  • (A) 6.02 × 1023 molecules
  • (B) 1.20 × 1024 molecules
  • (C) 3.01 × 1023 molecules
  • (D) 1.80 × 1024 molecules
View Answer
Explanation: Stoichiometry gives 1.20 × 1024 molecules.
Correct Answer: (B)

Question 3

Percent Yield Calculation: In the synthesis of ammonia (NH3), the theoretical yield is 45 g. If 30 g of ammonia are actually produced, what is the percent yield?

  • (A) 66.7%
  • (B) 75%
  • (C) 88.9%
  • (D) 50%
View Answer
Explanation: Percent yield = (Actual/Theoretical) × 100 = 66.7%.
Correct Answer: (A)

Question 4

Combustion of Hydrocarbon: A 9.0 g hydrocarbon produces 11.0 g of CO2. What is the mass percent of carbon in the hydrocarbon?

  • (A) 27.3%
  • (B) 36.4%
  • (C) 45.5%
  • (D) 72.7%
View Answer
Explanation: Mass percent of carbon = 36.4%.
Correct Answer: (B)

Question 5

Glass Composition: A glass composition contains 12% Na2O, 12% CaO, and 76% SiO2. Arrange them in order of moles from the most to the least.

  • (A) SiO2, CaO, Na2O
  • (B) SiO2, Na2O, CaO
  • (C) Na2O, CaO, SiO2
  • (D) Na2O, SiO2, CaO
View Answer
Explanation: SiO2 has the most moles.
Correct Answer: (A)

Question 6

Hydrate of Copper Sulfate: A hydrate of CuSO4 is heated, and its mass decreases from 125 g to 80 g. What is the formula for the hydrate?

  • (A) CuSO4·10H2O
  • (B) CuSO4·5H2O
  • (C) CuSO4·2H2O
  • (D) CuSO4·H2O
View Answer
Explanation: Mass loss corresponds to water content. The correct formula is CuSO4·5H2O.
Correct Answer: (B)

Question 7

Empirical Formula of Sulfur-Oxygen Compound: A sulfur-oxygen compound is 50% oxygen by weight. What is its empirical formula?

  • (A) SO
  • (B) SO2
  • (C) S2O3
  • (D) S2O5
View Answer
Explanation: The empirical formula is found to be SO2.
Correct Answer: (B)

Question 8

Sodium and Water Reaction: Sodium reacts with water to form 10 L of H2 at STP. What is the mass of sodium?

  • (A) 5 g
  • (B) 10 g
  • (C) 20 g
  • (D) 40 g
View Answer
Explanation: Using stoichiometry, the mass of sodium is calculated to be 20 g.
Correct Answer: (C)

Question 9

Atomic Mass of Element X: Four compounds containing element X have masses of 36 g, 54 g, 72 g, and 108 g. What is the possible atomic mass of X?

  • (A) 18.0
  • (B) 25.0
  • (C) 72.0
  • (D) 108
View Answer
Explanation: The masses suggest the possible atomic mass of X is 18.0.
Correct Answer: (A)

Question 10

Equimolar Solution Mixing: If equimolar solutions of NaCl and Pb(NO3)2 are mixed, which ion will not be present in significant amounts in the resulting solution after the reaction is completed?

  • (A) Na+
  • (B) Cl
  • (C) Pb2+
  • (D) NO3
View Answer
Explanation: PbCl2 precipitates, leaving Na+ and NO3.
Correct Answer: (B)

Question 11

Net Ionic Equation: Choose the correct net ionic equation representing the reaction that occurs between a solution of ammonium carbonate and a solution of copper (I) chloride.

  • (A) 2Cu+(aq) + CO32−(aq) → Cu2CO3(s)
  • (B) Cu2+(aq) + CO32−(aq) → CuCO3(s)
  • (C) (NH4)2CO3(aq) + 2CuCl(aq) → 2NH4Cl(aq) + Cu2CO3(s)
  • (D) (NH4)2CO3(aq) + 2CuCl(aq) → 2NH4Cl(s) + Cu2CO3(aq)
View Answer
Explanation: The net ionic equation involves the formation of Cu2CO3(s).
Correct Answer: (A)

Question 12

Oxidation State: Which of the following statements about the reaction given below is NOT true?
2Cr3+ + 3I2 + 7H2O → Cr2O72− + 6I + 14H+

  • (A) The oxidation number of chromium changes from +3 to +6.
  • (B) The oxidation number of iodine changes from 0 to -1.
  • (C) The oxidation number of hydrogen changes from +1 to 0.
  • (D) The oxidation number of oxygen remains the same.
View Answer
Explanation: Hydrogen’s oxidation number remains +1; it does not change.
Correct Answer: (C)

Question 13

Precipitate Formation: When aqueous solutions of iron (III) nitrate and sodium carbonate are mixed, what is the formula of the precipitate?

  • (A) Fe3(CO3)3
  • (B) Fe2(CO3)3
  • (C) NaNO3
  • (D) No precipitate would form.
View Answer
Explanation: Iron (III) carbonate, Fe2(CO3)3, precipitates out of solution.
Correct Answer: (B)

Question 14

Spectator Ions: If solutions containing AgNO3 and KBr are mixed, what are the spectator ions?

  • (A) Ag+ and Br
  • (B) K+ and Ag+
  • (C) K+ and NO3
  • (D) Ag+, NO3, K+, and Br
View Answer
Explanation: The spectator ions are K+ and NO3.
Correct Answer: (C)

Question 15

Half-Reaction Coefficient: …CN + …OH → …CNO + …H2O + …e
When the half-reaction above is balanced, what is the coefficient for e if all the coefficients are reduced to the lowest whole number?

  • (A) 1
  • (B) 2
  • (C) 3
  • (D) 4
View Answer
Explanation: Balancing the reaction gives a coefficient of 1 for electrons.
Correct Answer: (A)

Question 16

Decomposition of Potassium Chlorate: When heated, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

If 12.25 g of potassium chlorate decomposes completely, how many grams of oxygen gas are produced?

  • (A) 3.00 g
  • (B) 4.80 g
  • (C) 6.00 g
  • (D) 7.20 g
View Answer
Explanation: Using stoichiometry, 12.25 g of KClO3 produces 4.80 g of O2.
Correct Answer: (B)

Question 17

Volume of Oxygen Gas: Approximately how many liters of oxygen gas will be evolved at STP when 12.25 g of potassium chlorate decomposes according to the reaction:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

  • (A) 2.24 L
  • (B) 3.36 L
  • (C) 4.48 L
  • (D) 22.4 L
View Answer
Explanation: At STP, 12.25 g of KClO3 produces approximately 3.36 L of O2.
Correct Answer: (B)

Question 18

Effect of Temperature on Gas Pressure: If the temperature of the gas is doubled while the volume is held constant, what will happen to the pressure exerted by the gas and why?

  • (A) It will also double, because the gas molecules will be moving faster.
  • (B) It will also double, because the gas molecules are exerting a greater force on each other.
  • (C) It will be cut in half, because the molecules will lose more energy when colliding.
  • (D) It will increase by a factor of 4, because the kinetic energy will be four times greater.
View Answer
Explanation: According to the ideal gas law, doubling the temperature doubles the pressure.
Correct Answer: (A)

Question 19

Solution Concentration: How many grams of sodium chloride (NaCl) are needed to prepare 500 mL of a 0.5 M NaCl solution?

  • (A) 14.6 g
  • (B) 29.2 g
  • (C) 58.4 g
  • (D) 116.8 g
View Answer
Explanation: 0.25 moles of NaCl are required, which equals 14.6 g.
Correct Answer: (A)

Question 20

Gas Stoichiometry at Non-STP Conditions: What volume will 2.0 moles of an ideal gas occupy at a pressure of 2.0 atm and a temperature of 273 K?

  • (A) 22.4 L
  • (B) 11.2 L
  • (C) 24.0 L
  • (D) 44.8 L
View Answer
Explanation: Using PV = nRT, the gas occupies 22.4 L.
Correct Answer: (A)

Question 21

Gas Laws Application: A gas occupies 5.0 liters at a pressure of 1.0 atm. What will be its volume if the pressure is increased to 2.5 atm while the temperature remains constant?

  • (A) 2.0 L
  • (B) 3.0 L
  • (C) 5.0 L
  • (D) 12.5 L
View Answer
Explanation: Using Boyle’s law, the new volume is 2.0 L.
Correct Answer: (A)

Question 22

Ideal Gas Law Calculation: How many moles of gas are present in a 10.0 L container at 2.0 atm pressure and 300 K temperature?

  • (A) 0.80 mol
  • (B) 0.50 mol
  • (C) 1.00 mol
  • (D) 2.00 mol
View Answer
Explanation: Using PV = nRT, the number of moles is 0.80 mol.
Correct Answer: (A)

Question 23

Precipitate Formation: If solutions of barium nitrate and sodium sulfate are mixed, what will be the formula of the precipitate?

  • (A) BaSO4
  • (B) NaNO3
  • (C) Ba(NO3)2
  • (D) No precipitate will form.
View Answer
Explanation: BaSO4 precipitates as it is insoluble in water.
Correct Answer: (A)

Question 24

Spectator Ions: If solutions containing Na2SO4 and CaCl2 are mixed, which ions are the spectator ions in the resulting solution?

  • (A) Na+ and Cl
  • (B) Ca2+ and SO42−
  • (C) Na+ and SO42−
  • (D) Ca2+ and Cl
View Answer
Explanation: Spectator ions are Na+ and Cl.
Correct Answer: (A)

Question 25

Equilibrium of Ions: If equimolar solutions of Na2CO3 and Mg(NO3)2 are mixed, which ion will NOT be present in significant amounts in the resulting solution after equilibrium is established?

  • (A) Na+
  • (B) CO32−
  • (C) NO3
  • (D) Mg2+
View Answer
Explanation: Mg2+ reacts with CO32− to form insoluble MgCO3.
Correct Answer: (B)

Question 26

Net Ionic Equation: Choose the correct net ionic equation representing the reaction that occurs when solutions of silver nitrate and sodium carbonate are mixed.

  • (A) AgNO3(aq) + Na2CO3(aq) → NaNO3(aq) + Ag2CO3(s)
  • (B) 2Ag+(aq) + CO32−(aq) → Ag2CO3(s)
  • (C) Ag+(aq) + CO32−(aq) → Ag2CO3(aq)
  • (D) 2Ag+(aq) + 2NO3 → AgNO3(s)
View Answer
Explanation: Ag+ and CO32− form insoluble Ag2CO3.
Correct Answer: (B)

Question 27

Stoichiometry of Combustion: When 8.0 grams of methane (CH4) is combusted in excess oxygen, how many grams of carbon dioxide (CO2) will be produced?

  • (A) 22.0 g
  • (B) 44.0 g
  • (C) 88.0 g
  • (D) 16.0 g
View Answer
Explanation: 8.0 g of CH4 produces 22.0 g of CO2.
Correct Answer: (A)

Question 28

Stoichiometry of Acid-Base Reaction: How many grams of sodium hydroxide (NaOH) are needed to completely neutralize 50.0 mL of 0.5 M sulfuric acid (H2SO4)?

  • (A) 2.0 g
  • (B) 4.0 g
  • (C) 8.0 g
  • (D) 1.0 g
View Answer
Explanation: 2.0 g of NaOH is required for neutralization.
Correct Answer: (A)

Question 29

Limiting Reagent: If 14.0 grams of nitrogen (N2) react with 6.0 grams of hydrogen (H2) to form ammonia (NH3), what is the limiting reagent and how many grams of ammonia will be produced?

  • (A) N2; 17.0 g
  • (B) H2; 34.0 g
  • (C) N2; 34.0 g
  • (D) H2; 17.0 g
View Answer
Explanation: N2 is the limiting reagent, producing 17.0 g of NH3.
Correct Answer: (A)

Question 30

Gas Stoichiometry: How many liters of hydrogen gas at STP are produced when 10.0 grams of zinc react with excess hydrochloric acid according to the reaction:

Zn + 2HCl → ZnCl2 + H2

  • (A) 3.42 L
  • (B) 4.48 L
  • (C) 7.85 L
  • (D) 11.2 L
View Answer
Explanation: 10.0 g of Zn produces 3.42 L of H2.
Correct Answer: (A)

Question 31

Percent Yield Calculation: When 50.0 grams of calcium carbonate (CaCO3) is decomposed, it produces 20.0 grams of calcium oxide (CaO). If the theoretical yield is 28.0 grams, what is the percent yield of the reaction?

  • (A) 50%
  • (B) 71.4%
  • (C) 75%
  • (D) 85.7%
View Answer
Explanation: Percent yield = 71.4%.
Correct Answer: (B)

Question 32

Percent Error Calculation: During an experiment, 0.85 moles of hydrogen gas was collected. The theoretical amount was calculated to be 1.00 mole. What is the percent error of the experiment?

  • (A) 8.5%
  • (B) 10.0%
  • (C) 15.0%
  • (D) 20.0%
View Answer
Explanation: Percent error = |(1.00 – 0.85) / 1.00| × 100 = 15.0%.
Correct Answer: (C)

Question 33

Limiting Reagent and Stoichiometry: If 10.0 grams of sulfuric acid (H2SO4) reacts with 10.0 grams of sodium hydroxide (NaOH), what mass of sodium sulfate (Na2SO4) will be formed?

  • (A) 12.5 g
  • (B) 14.2 g
  • (C) 17.5 g
  • (D) 20.0 g
View Answer
Explanation: Limiting reagent is H2SO4, producing 14.2 g of Na2SO4.
Correct Answer: (B)

Question 34

Gas Stoichiometry: How many liters of carbon dioxide (CO2) gas at STP will be produced when 5.0 grams of calcium carbonate (CaCO3) decomposes?

  • (A) 1.12 L
  • (B) 2.24 L
  • (C) 3.36 L
  • (D) 4.48 L
View Answer
Explanation: 5.0 g of CaCO3 produces 1.12 L of CO2.
Correct Answer: (A)

Question 35

Mass-to-Mass Stoichiometry: How many grams of water (H2O) will be produced when 10.0 grams of hydrogen gas (H2) reacts completely with oxygen gas (O2)?

  • (A) 90.0 g
  • (B) 80.0 g
  • (C) 100.0 g
  • (D) 180.0 g
View Answer
Explanation: 10.0 g of H2 produces 90.0 g of H2O.
Correct Answer: (A)

Question 36

Excess Reagent Calculation: If 20.0 grams of magnesium reacts with 10.0 grams of hydrochloric acid (HCl), what is the mass of excess magnesium left after the reaction?

  • (A) 5.2 g
  • (B) 9.8 g
  • (C) 10.0 g
  • (D) 15.0 g
View Answer
Explanation: Excess Mg = 9.8 g after the reaction.
Correct Answer: (B)

Question 37

Limiting Reagent and Percent Yield: When 15.0 grams of aluminum reacts with 50.0 grams of iron(III) oxide (Fe2O3), how many grams of iron (Fe) will be produced if the reaction has a 75% yield? The balanced equation for the reaction is: 2Al + Fe2O3 → Al2O3 + 2Fe.

  • (A) 20.5 g
  • (B) 30.0 g
  • (C) 35.5 g
  • (D) 40.0 g
View Answer
Explanation: Actual Fe = 20.5 g with 75% yield.
Correct Answer: (A)

Question 38

Combustion Analysis: A 10.0 g sample of a hydrocarbon is completely burned in oxygen, producing 33.0 g of CO2 and 13.5 g of H2O. Determine the empirical formula of the hydrocarbon.

  • (A) CH
  • (B) C2H4
  • (C) C3H6
  • (D) C4H8
View Answer
Explanation: Empirical formula = CH2.
Correct Answer: (B)

Question 39

Gas Collection Over Water: In an experiment, 2.00 grams of zinc reacts with excess hydrochloric acid to produce hydrogen gas. The gas is collected over water at 25°C, where the vapor pressure of water is 24 mmHg. If the total pressure in the container is 780 mmHg, what is the volume of the dry hydrogen gas collected at STP?

  • (A) 0.745 L
  • (B) 0.815 L
  • (C) 0.930 L
  • (D) 1.10 L
View Answer
Explanation: Volume of dry H2 gas = 0.745 L at STP.
Correct Answer: (A)

Question 40

Back Titration: A sample of impure limestone weighing 2.50 grams is dissolved in 50.0 mL of 1.0 M hydrochloric acid (HCl). The excess acid is then back-titrated with 0.50 M sodium hydroxide (NaOH), requiring 30.0 mL to reach the endpoint. Calculate the percentage purity of the limestone (assumed to be CaCO3).

  • (A) 65.0%
  • (B) 70.0%
  • (C) 75.0%
  • (D) 85.0%
View Answer
Explanation: Percentage purity = 70.0%.
Correct Answer: (B)

Question 41

Equilibrium Constant: At a certain temperature, the equilibrium constant for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is Keq = 0.500. If [N2] = 0.200 M, [H2] = 0.300 M, and [NH3] = 0.400 M, is the system at equilibrium?

  • (A) Yes
  • (B) No, the reaction will shift left.
  • (C) No, the reaction will shift right.
  • (D) Cannot be determined.
View Answer
Explanation: The reaction will shift left to reduce [NH3].
Correct Answer: (B)

Question 42

Stoichiometry and Mass Conservation: A 10.0 g sample of calcium carbonate (CaCO3) is heated strongly until it decomposes completely into calcium oxide (CaO) and carbon dioxide (CO2), according to the reaction:

CaCO3(s) → CaO(s) + CO2(g)

Calculate the mass of calcium oxide produced.

  • (A) 5.60 g
  • (B) 8.40 g
  • (C) 10.0 g
  • (D) 4.40 g
View Answer

Answer Explanation:

Molar mass of CaCO3 = 100.1 g/mol. Molar mass of CaO = 56.1 g/mol.

Moles of CaCO3 = 10.0 g / 100.1 g/mol = 0.10 moles.

From the balanced equation, 1 mole of CaCO3 produces 1 mole of CaO.

Mass of CaO = 0.10 moles × 56.1 g/mol = 5.60 g.

Correct Answer: (A)


Question 43

Stoichiometry: In a reaction between 5.00 g of aluminum (Al) and 20.00 g of chlorine gas (Cl2), what is the limiting reagent? How many grams of aluminum chloride (AlCl3) are produced?

  • (A) Al is the limiting reagent; 13.33 g AlCl3 produced.
  • (B) Cl2 is the limiting reagent; 26.67 g AlCl3 produced.
  • (C) Al is the limiting reagent; 26.67 g AlCl3 produced.
  • (D) Cl2 is the limiting reagent; 13.33 g AlCl3 produced.
View Answer
Explanation: Use stoichiometry with the balanced equation: 2Al + 3Cl2 → 2AlCl3.
Correct Answer: (D)

Question 44

Stoichiometry in Gas Reactions: When 10.0 grams of propane (C3H8) is combusted in excess oxygen, how many liters of carbon dioxide (CO2) are produced at STP?

The balanced equation for the combustion reaction is:

C3H8 + 5O2 → 3CO2 + 4H2O

  • (A) 22.4 L
  • (B) 33.6 L
  • (C) 44.8 L
  • (D) 11.2 L
View Answer

Answer Explanation:

Molar mass of C3H8 = 3(12.01) + 8(1.008) = 44.09 g/mol.

Moles of C3H8 = 10.0 g / 44.09 g/mol ≈ 0.227 moles.

From the balanced equation, 1 mole of C3H8 produces 3 moles of CO2.

Moles of CO2 = 0.227 moles × 3 = 0.681 moles.

At STP, 1 mole of gas occupies 22.4 L. Volume of CO2 = 0.681 moles × 22.4 L/mol ≈ 15.26 L.

Correct Answer: (B)


Question 45

Limiting Reagent and Excess: 8.00 g of sodium reacts with 10.00 g of chlorine gas to form sodium chloride. Identify the limiting reagent and calculate the mass of excess reagent remaining after the reaction.

  • (A) Na is the limiting reagent; 4.45 g Cl2 remains.
  • (B) Cl2 is the limiting reagent; 3.55 g Na remains.
  • (C) Na is the limiting reagent; 3.55 g Cl2 remains.
  • (D) Cl2 is the limiting reagent; 4.45 g Na remains.
View Answer
Explanation: Use molar masses and stoichiometry to determine limiting reagent.
Correct Answer: (A)

Question 46

Stoichiometry and Percent Yield: If 25.00 g of iron(III) oxide (Fe2O3) reacts with excess carbon monoxide (CO) to produce iron (Fe) and carbon dioxide (CO2), and the reaction yield is 80.0%, how many grams of iron are produced?

  • (A) 14.93 g
  • (B) 20.90 g
  • (C) 29.85 g
  • (D) 40.00 g
View Answer
Explanation: Use the balanced equation Fe2O3 + 3CO → 2Fe + 3CO2 and apply the percent yield formula.
Correct Answer: (A)

Question 47

Reaction Stoichiometry: Ammonium nitrate (NH4NO3) decomposes to form nitrogen gas (N2), water (H2O), and oxygen gas (O2). If 16.0 g of NH4NO3 decomposes, what volume of oxygen gas is produced at STP?

  • (A) 5.60 L
  • (B) 8.96 L
  • (C) 11.20 L
  • (D) 22.40 L
View Answer
Explanation: Use the balanced equation and molar volume at STP (22.4 L/mol).
Correct Answer: (A)

Question 48

Stoichiometry in Solution: A solution of 50.0 mL of 0.500 M sulfuric acid (H2SO4) is neutralized by a solution of 0.250 M sodium hydroxide (NaOH). Calculate the volume of NaOH required to completely neutralize the acid.

  • (A) 50.0 mL
  • (B) 100.0 mL
  • (C) 200.0 mL
  • (D) 25.0 mL
View Answer

Answer Explanation:

The balanced equation is: H2SO4 + 2NaOH → Na2SO4 + 2H2O.

Moles of H2SO4 = 0.500 M × 0.0500 L = 0.0250 moles.

From the equation, 1 mole of H2SO4 reacts with 2 moles of NaOH, so moles of NaOH needed = 0.0250 × 2 = 0.0500 moles.

Volume of NaOH = Moles / Concentration = 0.0500 moles / 0.250 M = 0.200 L = 200.0 mL.

Correct Answer: (C)


Question 49

Stoichiometry in Solutions: How many grams of NaCl are needed to prepare 250.0 mL of a 0.500 M solution?

  • (A) 7.31 g
  • (B) 5.85 g
  • (C) 11.70 g
  • (D) 2.93 g
View Answer
Explanation: Use Molarity (M = moles/L) and molar mass of NaCl.
Correct Answer: (A)

Question 50

Titration and Stoichiometry: What volume of 0.200 M HCl is required to completely neutralize 25.00 mL of 0.100 M NaOH?

  • (A) 10.00 mL
  • (B) 12.50 mL
  • (C) 25.00 mL
  • (D) 50.00 mL
View Answer
Explanation: Use the relationship M1V1 = M2V2.
Correct Answer: (A)

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