Chapter9ChemistryReview_Sheet

Chemistry Review Sheet: Stoichiometry (Chapter 9)

  • Focus on the reaction of butane (C4H10) with oxygen gas to form carbon dioxide (CO2) and water (H2O).

  • All calculations must include the work shown for clarity.

1. Balanced Chemical Equation

  • For the reaction:

    • C4H10 + O2 -> CO2 + H2O

    • The balanced equation is:

    • 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

2. Moles of Oxygen Needed

  • Calculation for the stoichiometric relationship:

    • From the balanced equation, 2 moles of C4H10 react with 13 moles of O2.

    • To find moles of O2 needed for 3.6 moles of butane:

    • (3.6 moles C4H10) x (13 moles O2 / 2 moles C4H10) = 23.4 moles O2

3. Grams of CO2 Produced

  • To determine grams of CO2 from 1.25 moles of O2:

    • From the balanced equation, 13 moles of O2 produce 8 moles of CO2.

    • First, find moles of CO2:

      • (1.25 moles O2) x (8 moles CO2 / 13 moles O2) = 0.7692 moles CO2

    • Convert moles to grams:

      • 0.7692 moles CO2 x 44.01 g/mol = 33.88 g CO2

4. Moles of Butane from Grams of H2O

  • Calculation based on produced grams of H2O:

    • 17 g H2O to moles:

      • 17 g / 18.015 g/mol = 0.944 moles H2O

    • Find moles of butane reacting:

      • From balanced equation, 10 moles H2O produced from 2 moles C4H10.

      • (0.944 moles H2O) x (2 moles C4H10 / 10 moles H2O) = 0.1888 moles C4H10

5. Mass of Water Produced from Butane

  • If 55.85 g of butane reacts:

    • Convert grams of butane to moles:

      • 55.85 g C4H10 / 58.12 g/mol = 0.960 moles C4H10

    • Calculate moles of H2O produced:

      • (0.960 moles C4H10) x (10 moles H2O / 2 moles C4H10) = 4.8 moles H2O

    • Convert moles H2O to grams:

      • 4.8 moles H2O x 18.015 g/mol = 86.48 g H2O

6. Limiting Reactant

  • To identify the limiting reactant:

    • Compare available moles of butane and oxygen:

    • Use moles from previous calculations based on the balanced equation, the limiting reactant is the one that produces the least amount of product.

    • It limits the overall reaction and determines the amount of products formed.

7. Theoretical Yield of CO2

  • Calculation for 44.04 g butane mixed with 82.235 g O2:

    • First, convert grams to moles:

      • 44.04 g C4H10 / 58.12 g/mol = 0.7582 moles C4H10

      • 82.235 g O2 / 32.00 g/mol = 2.5772 moles O2

    • Determine the limiting reactant:

      • Required O2 for 0.7582 moles C4H10:

      • (0.7582 moles C4H10) x (13 moles O2 / 2 moles C4H10) = 4.9103 moles O2 needed

      • Here, O2 is in excess, so C4H10 is limiting. Calculate the moles of CO2 produced:

      • (0.7582 moles C4H10) x (8 moles CO2 / 2 moles C4H10) = 3.0328 moles CO2

      • Convert moles of CO2 to grams:

      • 3.0328 moles CO2 x 44.01 g/mol = 133.44 g CO2

8. Remaining Excess Reactant

  • For the reactant in excess (O2) after the reaction:

    • Calculate moles of all O2 used based on C4H10 reacted:

    • Moles of O2 used = (0.7582 moles C4H10) x (13 moles O2 / 2 moles C4H10) = 4.9103 moles O2

    • Remaining O2: 2.5772 moles available - 4.9103 moles used = -2.3331 moles (not enough O2)

9. % Yield of CO2 Produced

  • If 40.814 g CO2 is produced:

    • Calculate percent yield:

      • % Yield = (actual yield / theoretical yield) x 100

      • % Yield = (40.814 g / 133.44 g) x 100 = 30.6%

Chemistry Review Sheet: Stoichiometry (Chapter 9)

This review sheet focuses on the stoichiometric relationship in the reaction of butane (C4H10) with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O), fundamental in understanding combustion reactions and balancing chemical equations. Detailed calculations with work shown for clarity are integral to mastering stoichiometry.

  1. Balanced Chemical EquationFor the combustion reaction of butane:C4H10 + O2 -> CO2 + H2OThe balanced equation illustrates that:2 C4H10 + 13 O2 -> 8 CO2 + 10 H2OThis indicates that two moles of butane react with thirteen moles of oxygen to produce eight moles of carbon dioxide and ten moles of water.

  2. Moles of Oxygen NeededCalculating the stoichiometric relationship is crucial. From the balanced equation, 2 moles of C4H10 react with 13 moles of O2. To find the moles of O2 needed for 3.6 moles of butane:(3.6 moles C4H10) x (13 moles O2 / 2 moles C4H10) = 23.4 moles O2 needed. This calculation illustrates the direct proportionality in the reaction.

  3. Grams of CO2 ProducedTo determine the grams of CO2 produced from 1.25 moles of O2, we use the relationships established in the balanced equation.Given that 13 moles of O2 yield 8 moles of CO2, we first find moles of CO2:(1.25 moles O2) x (8 moles CO2 / 13 moles O2) = 0.7692 moles CO2.Next, converting moles to grams:0.7692 moles CO2 x 44.01 g/mol = 33.88 g CO2.This process exemplifies the conversion from moles to mass, significant in stoichiometric calculations.

  4. Moles of Butane from Grams of H2OCalculations based on produced grams of H2O allow us to determine the moles of butane used. To convert 17 g of H2O to moles:17 g / 18.015 g/mol = 0.944 moles H2O.Now, determining moles of butane that reacted:From the balanced equation, 10 moles of H2O are produced from 2 moles of C4H10:(0.944 moles H2O) x (2 moles C4H10 / 10 moles H2O) = 0.1888 moles C4H10.This calculation showcases how to derive the initial reactant amount from a product measurement.

  5. Mass of Water Produced from ButaneIn determining the mass of water formed from a specific amount of butane, say 55.85 g of C4H10:Convert grams of butane to moles:55.85 g C4H10 / 58.12 g/mol = 0.960 moles C4H10.Calculate the moles of H2O produced:(0.960 moles C4H10) x (10 moles H2O / 2 moles C4H10) = 4.8 moles H2O.Now, converting moles of H2O to grams:4.8 moles H2O x 18.015 g/mol = 86.48 g H2O.Understanding these conversions is paramount in quantitative chemistry.

  6. Limiting ReactantTo identify the limiting reactant, compare the available moles of butane and oxygen. By utilizing the moles from previous calculations based on the balanced equation, the limiting reactant is defined as the reactant that will produce the least amount of product, thus limiting the extent of the reaction.

  7. Theoretical Yield of CO2For a practical example, calculating the theoretical yield of CO2 when 44.04 g of butane is mixed with 82.235 g of O2:Firstly, convert grams to moles:44.04 g C4H10 / 58.12 g/mol = 0.7582 moles C4H10, 82.235 g O2 / 32.00 g/mol = 2.5772 moles O2.Next, determine the limiting reactant by calculating the O2 required for the available butane:(0.7582 moles C4H10) x (13 moles O2 / 2 moles C4H10) = 4.9103 moles O2 needed.Since the available O2 is in excess, C4H10 is the limiting reactant, giving us:(0.7582 moles C4H10) x (8 moles CO2 / 2 moles C4H10) = 3.0328 moles CO2 produced.Finally, converting moles of CO2 to grams gives:3.0328 moles CO2 x 44.01 g/mol = 133.44 g CO2.Such calculations help predict product formation in chemical reactions effectively.

  8. Remaining Excess ReactantTo find the remaining reactant in excess (O2) after the reaction, we calculate how much O2 is used based on the amount of C4H10 that reacted:Moles of O2 used = (0.7582 moles C4H10) x (13 moles O2 / 2 moles C4H10) = 4.9103 moles O2.The remaining amount of O2 is:2.5772 moles available - 4.9103 moles used = -2.3331 moles (indicating a deficit).Understanding excess reactants is key to optimizing reactant usage in reactions.

  9. % Yield of CO2 ProducedIf an actual yield of CO2 produced is 40.814 g, we calculate the percent yield as follows: % Yield = (actual yield / theoretical yield) x 100% Yield = (40.814 g / 133.44 g) x 100 = 30.6%.The percent yield is a crucial indicator for evaluating the efficiency and success of a reaction in practical applications.