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Sample Quiz: Chemistry Chapter 3

Chemical Reactions and Stoichiometry — Chemistry: The Central Science, 15th Global Edition in SI Units (Theodore L. Brown, H. Eugene LeMay, Jr., Bruce E. Bursten, Catherine J. Murphy, Patrick M. Woodward, Matthew W. Stoltzfus)

10 questions · 23 points · answer key included

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Multiple Choice

  1. ApplyWhen balancing a chemical equation, what is the purpose of using coefficients rather than changing subscripts? (2 PTS)
    1. Subscripts cannot be used in chemical equations
    2. Coefficients make the equation easier to read
    3. Coefficients change the amount of substance without changing the identity of the compound
    4. Coefficients are always smaller numbers than subscripts
  2. ApplyIn the combustion of methane, CH4 + 2O2 arrow CO2 + 2H2O, how many total atoms are on the left side of the equation? (2 PTS)
    1. 9 atoms
    2. 15 atoms
    3. 5 atoms
    4. 12 atoms
  3. ApplyWhich of the following represents a combination reaction? (2 PTS)
    1. CaCO3(s) arrow CaO(s) + CO2(g)
    2. 2NaN3(s) arrow 2Na(s) + 3N2(g)
    3. 2Mg(s) + O2(g) arrow 2MgO(s)
    4. CH3OH(l) + O2(g) arrow CO2(g) + H2O(g)
  4. ApplyWhen magnesium metal reacts with oxygen gas to form magnesium oxide, what type of compound is produced? (2 PTS)
    1. A gaseous compound
    2. A liquid solution
    3. A covalent molecule
    4. An ionic solid
  5. ApplyIn the decomposition of sodium azide used in airbags, 2NaN3(s) arrow 2Na(s) + 3N2(g), approximately how much nitrogen gas is produced from 100 g of NaN3? (2 PTS)
    1. About 25 L of N2
    2. About 100 L of N2
    3. About 75 L of N2
    4. About 50 L of N2

True or False

  1. In a decomposition reaction, a single substance breaks down to produce two or more products. (1 PTS)
  2. The molar mass of a substance in grams per mole is numerically equal to its formula weight expressed in atomic mass units. (1 PTS)
  3. When balancing a chemical equation, you may change the subscripts in a chemical formula to balance the number of atoms on each side. (1 PTS)

Problem Solving

  1. ApplyCalculate the number of moles of glucose (C6H12O6) in a 5.380 g sample. (Atomic weights: C = 12.0 amu, H = 1.0 amu, O = 16.0 amu) (5 PTS)
  2. ApplyBalance the following equation and identify the type of reaction: C7H16(s) + O2(g) arrow CO2(g) + H2O(l) (5 PTS)

Answer key

Multiple Choice

  1. 1.C — Coefficients adjust the relative amounts of reactants and products while preserving the chemical identity, whereas changing subscripts would create different compounds entirely.
  2. 2.B — The left side has 1 CH4 molecule (5 atoms: 1 C and 4 H) plus 2 O2 molecules (4 atoms total), giving 5 + 4 = 9 atoms; however, counting all atoms: 1 C + 4 H + 4 O = 9 atoms on left, but the question asks for total atoms which is 1 + 4 + 4 = 9. Rechecking: CH4 has 5 atoms, 2O2 has 4 atoms, total = 9 atoms on left side.
  3. 3.C — A combination reaction involves two or more substances reacting to form one product; 2Mg(s) + O2(g) arrow 2MgO(s) fits this pattern with two reactants forming one product.
  4. 4.D — The reaction between a metal (Mg) and a nonmetal (O) produces an ionic solid, as Mg loses electrons to form Mg2+ and O gains electrons to form O2-.
  5. 5.D — According to the textbook, approximately 100 g of NaN3 will explosively produce about 50 L of nitrogen gas.

True or False

  1. 1.True — The textbook defines decomposition reactions as reactions in which a single substance undergoes a reaction to produce two or more products, as shown by the general form C → A + B.
  2. 2.True — The textbook explicitly states that the molar mass in grams per mole of any substance is numerically equal to its formula weight in atomic mass units, with the example that H₂O has a molecular weight of 18.0 amu and a molar mass of 18.0 g/mol.
  3. 3.False — The textbook clearly states that changing subscripts in a formula changes the identity of the substance and should never be done when balancing an equation; only coefficients should be adjusted.

Problem Solving

  1. 1.First, calculate the molar mass of glucose: M = 6(12.0) + 12(1.0) + 6(16.0) = 72.0 + 12.0 + 96.0 = 180.0 g/mol. Then use dimensional analysis: moles = 5.380 g × (1 mol)/(180.0 g) = 0.02989 mol ≈ 0.0299 mol (or 2.99 × 10-2 mol with 3 significant figures). — This problem requires calculating molar mass from atomic weights and then converting grams to moles using the molar mass as a conversion factor, demonstrating the application of the mole concept.
  2. 2.First, balance carbon: 7 carbons on left require coefficient 7 for CO2. Next, balance hydrogen: 16 hydrogens on left require coefficient 8 for H2O (since each H2O has 2 H). Finally, balance oxygen: products have 7(2) + 8(1) = 22 oxygen atoms, requiring coefficient 11 for O2 (since each O2 has 2 O). Balanced equation: C7H16(s) + 11O2(g) arrow 7CO2(g) + 8H2O(l). This is a combustion reaction because a hydrocarbon reacts with oxygen to produce carbon dioxide and water. — This problem requires systematic balancing of a combustion equation and classification of the reaction type, applying the procedure for balancing equations and understanding reaction patterns.