Unit 4 Lesson 1: Stoichiometry Practice Questions

Introduction

  • This session focuses on the practice of stoichiometry through various reaction equations and calculations related to moles, mass, and molecules.
  • Essential to show all work in calculations to reinforce understanding of stoichiometric principles.

Question Set 1: Reaction of Gold and Oxygen

Chemical Equation
  • Unbalanced Reaction: ______ Au + ______ O2 → _______ Au2O3
Specific Problems

  1. a) Moles of Gold Needed

    • Question: How many moles of gold would be needed to react with 3.82 moles of oxygen?
    • Given: 3.82 moles O2
    • Stoichiometric Ratio: Mole ratio of Au to O2 is determined from the balanced equation (to be established).
    • Calculation: Let x = moles of Au needed.
    • If equation is balanced (2 Au + 3 O2 → 2 Au2O3), stoichiometric ratio = 2/3.
    • Moles of Au needed for 3.82 moles O2:
      x = 3.82 imes rac{2}{3} = 2.55 ext{ moles Au}

  2. b) Molecules of Gold (III) Oxide Produced

    • Question: How many molecules of gold (III) oxide can be produced from 13.5 moles of gold?
    • Given: 13.5 moles Au
    • Stoichiometric Ratio: Mole ratio of Au to Au2O3 is from the balanced equation (to be established).
    • Calculation: Using mole conversion (1 mol Au2O3 = 2 mol Au), we have:
      ext{Moles of } Au2O3 = rac{13.5}{2} = 6.75 ext{ mol}
    • Converting moles of Au2O3 to molecules:
      extMolecules=6.75imes(6.022imes1023)extmolecules/mol=4.06imes1024extmoleculesAu2O3ext{Molecules} = 6.75 imes (6.022 imes 10^{23}) ext{ molecules/mol} = 4.06 imes 10^{24} ext{ molecules Au2O3}

  3. c) Moles of O2 Needed to Produce Au2O3

    • Question: How many moles of O2 are needed to produce 34.7 g of Au2O3?
    • Given: 34.7 g of Au2O3
    • Molar Mass of Au2O3: (2197) + (316) = 233 g/mol
    • Moles of Au2O3:
      n = rac{34.7 ext{ g}}{233 ext{ g/mol}} = 0.149 ext{ mol Au2O3}
    • Stoichiometric Ratio: From balanced equation (2/3), moles of O2:
      n = 0.149 imes rac{3}{2} = 0.224 ext{ mol O2}

Question Set 2: Reaction of Butane (C4H10) with Oxygen

Chemical Equation
  • Unbalanced Reaction: ______ C4H10 + ______ O2 → ______ CO2 + ______ H2O
Specific Problems

  1. a) Moles of Carbon Dioxide Produced

    • Question: When 0.624 moles of O2 are reacted, how many moles of carbon dioxide are produced?
    • Given: 0.624 moles O2
    • Stoichiometric Ratio: Assuming balanced equation yields CO2 and water, C4H10 reacts with O2 (Balanced equation to be determined).
    • Calculation: Let’s establish if balanced (C4H10 + 13/2 O2 → 4 CO2 + 5 H2O) leads O2 to CO2 with ratio 1:4.
      extMolesCO2=0.624imes4=2.496extmolesCO2ext{Moles CO2} = 0.624 imes 4 = 2.496 ext{ moles CO2}
    • Simplified Answer: 0.384 mol CO2

  2. b) Grams of C4H10 Needed

    • Question: How many grams of C4H10 are needed to produce 3.7 moles of water?
    • Given: 3.7 moles H2O
    • Stoichiometric Ratio: Balanced equation indicates H2O to C4H10 ratio, typically 5 H2O per 1 C4H10 (assuming balanced).
    • Calculation:
      ext{Moles of C4H10} = rac{3.7}{5} = 0.74 ext{ moles C4H10}
    • Molar Mass of C4H10: (412) + (101) = 58 g/mol
    • Grams of C4H10:
      extMass=0.74imes58=42.92extgext{Mass} = 0.74 imes 58 = 42.92 ext{ g}
    • Simplified Answer: 43 g C4H10

  3. c) Volume of O2 Gas at STP

    • Question: What volume of O2 gas, at STP, is needed to react with 2.56 g of C4H10?
    • Molar Mass of C4H10: 58 g/mol
    • Moles of C4H10:
      n = rac{2.56}{58} = 0.044 g/mol
    • Stoichiometric Ratio: Using balanced equation’s oxygen to butane conversion ratio, airflow typical = 1 to 13/2 yields needed O2.
    • Calculate O2 moles needed, then convert to volume at STP ($22.4 rac{L}{mol}$):
      n_{O2} = 0.044 imes rac{13}{2} = 0.286 ext{ moles O2}
      V=nimes22.4=6.43extLO2V = n imes 22.4 = 6.43 ext{ L O2}
    • Simplified Answer: 6.43 L O2

Question Set 3: Decomposition of Potassium Chlorate

Chemical Equation
  • Unbalanced Reaction: ______ KClO3 → ______ KCl + ______ O2
Specific Problems

  1. a) Moles of KCl Formed

    • Question: When 62.0 g of Potassium chlorate are reacted, how many moles of KCl will be formed?
    • Given: 62.0 g KClO3
    • Molar Mass of KClO3: (39) + (35.5) + (3 imes 16) = 122.5 g/mol
    • Moles of KClO3:
      n = rac{62.0}{122.5} = 0.506 ext{ mol KClO3}
    • Stoichiometric Ratio: KClO3 decomposes to KCl yields 1:1 ratio (assumed from balanced).
      extMolesofKCl=n=0.506extmolKClext{Moles of KCl} = n = 0.506 ext{ mol KCl}
    • Simplified Answer: 0.506 mol KCl

  2. b) Molecules of O2 Produced

    • Question: How many molecules of O2 are produced from 2.85 g of KClO3?
    • Given: 2.85 g KClO3
    • Moles of KClO3:
      n = rac{2.85}{122.5} = 0.0232 ext{ mol KClO3}
    • Stoichiometric Ratio: Assuming decomposition yields O2 and KCl (1:1:0.5 basis).
      ext{Moles of O2} = 0.0232 imes rac{1}{2} = 0.0116 ext{ mol O2}
    • Converting moles to molecules:
      extMoleculesO2=0.0116imes(6.022imes1023)=2.10imes1022ext{Molecules O2} = 0.0116 imes (6.022 imes 10^{23}) = 2.10 imes 10^{22}
    • Simplified Answer: 2.10 x 10^22 molecules O2

  3. c) Grams of KClO3 for O2 Production

    • Question: 3.54 g of oxygen was produced. How many grams of potassium chlorate were used?
    • Given: 3.54 g O2
    • Molar Mass of O2: 32 g/mol
    • Moles of O2:
      n = rac{3.54}{32} = 0.1106 ext{ mol O2}
    • Stoichiometric Ratio: Assuming above derived mole relationships connected back to KClO3, since 2 mol KClO3 yield 1 mol O2.
      nKClO3=0.1106imes2=0.2212extmolKClO3n_{KClO3} = 0.1106 imes 2 = 0.2212 ext{ mol KClO3}
    • Mass of KClO3:
      extMass=nimesextMolarMass=0.2212imes122.5=27.1extgKClO3ext{Mass} = n imes ext{Molar Mass} = 0.2212 imes 122.5 = 27.1 ext{ g KClO3}
    • Simplified Answer: 9.04 g KClO3

Question Set 4: Reaction of Boron Trifluoride with Water

Chemical Equation
  • Unbalanced Reaction: ______ BF3 + ______ H2O → ______ B2O3 + ______ HF
Specific Problems

  1. a) Mass of Diboron Trioxide Produced

    • Question: 284 g of boron trifluoride can produce what mass (in kilograms) of diboron trioxide?
    • Given: 284 g BF3
    • Molar Mass of BF3: (10.81) + (3*19 = 57) = 67.81 g/mol
    • Moles of BF3:
      n = rac{284}{67.81} = 4.18 ext{ mol BF3}
    • Stoichiometric Ratio: Balanced equation indicates 2 BF3 yield 1 B2O3, ratio of 2:1.
      Moles B2O3 = 4.18 imes rac{1}{2} = 2.09 ext{ mol B2O3}
    • Molar Mass of B2O3: (210.81) + (316) = 69.62 g/mol
    • kg of B2O3:
      extMass=2.09imes69.62=145.6g=0.146extkgB2O3ext{Mass} = 2.09 imes 69.62 = 145.6 g = 0.146 ext{ kg B2O3}
    • Simplified Answer: 0.146 kg B2O3

  2. b) Mass of Water Reacted from Hydrofluoric Acid Produced

    • Question: If 0.835 L of hydrofluoric acid gas at STP are produced, what mass (in grams) of water was reacted?
    • Given: Volume of HF generated at STP = 0.835 L
    • Molar Volume at STP: 22.4 L/mol
    • Moles of HF:
      n = rac{0.835}{22.4} = 0.0372 ext{ mol HF}
    • Stoichiometric Ratio: From the original unbalanced equation, assume 2 HF produced for each 2 BF3.
    • Moles of water:
      nH2O=0.0372extmoln_{H2O} = 0.0372 ext{ mol}
    • Molar Mass of H2O: 18 g/mol
    • Mass of H2O reacted:
      extMass=0.0372imes18=0.335extgH2Oext{Mass} = 0.0372 imes 18 = 0.335 ext{ g H2O}
    • Simplified Answer: 0.335 g H2O

  3. c) Number of Oxygen Atoms from BF3

    • Question: How many oxygen atoms can be produced from 5.7 moles of BF3?

    • Given: 5.7 moles BF3

    • Stoichiometric Ratio: From balanced equation shows each 2 BF3 produces one molecule of B2O3 releasing 3 O atoms, hence scaling down each BF3.

    • Moles of produced oxygen atoms:
      moles = 5.7 imes rac{3}{2} = 8.55 ext{ moles O atoms}

    • Number of Oxygen Atoms:
      Ototal=8.55imes(6.022imes1023)extatoms/mole=5.1imes1024extoxygenatomsO_{total} = 8.55 imes (6.022 imes 10^{23}) ext{ atoms/mole} = 5.1 imes 10^{24} ext{ oxygen atoms}

    • Simplified Answer: 5.1 x 10^{24} oxygen atoms.