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CHAPTER 14

CHAPTER 14. Gases. 14.3 Stoichiometry and Gases. Draw me. Ideal gas law. Me too! Me too!. We can now solve stoichiometry puzzles involving gases. R = universal gas constant. We can now solve stoichiometry puzzles involving gases. with 1. solids 2. solutions 3. other gases.

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CHAPTER 14

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  1. CHAPTER 14 Gases 14.3 Stoichiometry and Gases

  2. Draw me. Ideal gas law Me too! Me too! We can now solve stoichiometry puzzles involving gases R = universal gas constant

  3. We can now solve stoichiometry puzzles involving gases with 1. solids 2. solutions 3. other gases

  4. Steps for solving stoichiometry problems

  5. Steps for solving stoichiometry problems

  6. Draw me

  7. Ideal gas law We can now solve stoichiometry problems involving gases and solids solutions other gases R = universal gas law

  8. Stoichiometry Puzzle #1 of 3 Limiting reactant is a solid: If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400oC. Assuming the carbon dioxide produced is at that temperature, what volume of CO2 is produced at a pressure of 790 mmHg? The combustion reaction is: 2C20H42 + 61O2→ 40CO2 + 42H2O

  9. If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400oC. Assuming the carbon dioxide produced is at that temperature, what volume of CO2 is produced at a pressure of 790 mmHg? The combustion reaction is: 2C20H42 + 61O2→ 40CO2 + 42H2O Asked:Volume of CO2 produced Given: Paraffin: mass of 125 g CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT

  10. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 125 g C20H42

  11. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 0.442 moles C20H42 125 g C20H42

  12. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 0.442 moles C20H42

  13. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 0.442 moles C20H42 8.85 moles CO2

  14. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 8.85 moles CO2

  15. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 8.85 moles CO2

  16. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 8.85 moles CO2

  17. Asked: Volume of CO2 produced Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC Relationships: Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C20H42 ~ 40 moles CO2 PV = nRT Solve: 8.85 moles CO2 1,170 L CO2

  18. Ideal gas law We can now solve stoichiometry problems involving gases and solids solutions other gases R = universal gas law

  19. Stoichiometry Puzzle #2 of 3 Limiting reactant is a solution: A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium: If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22oC? Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)

  20. A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium: If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22oC? Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq) Asked:Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT

  21. Volume Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.050 L HCl 1.25 M HCl

  22. Volume Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.050 L HCl 1.25 M HCl 0.0625 moles HCl

  23. Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.0625 moles HCl

  24. Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.0625 moles HCl 0.0313 moles H2

  25. Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.0313 moles H2

  26. Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.0313 moles H2

  27. Asked: Pressure of H2 produced Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L H2: V = 1.00 L, T = 22oC Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2 PV = nRT Solve: 0.0313 moles H2 0.756 atm H2

  28. Ideal gas law We can now solve stoichiometry problems involving gases and solids solutions other gases R = universal gas law

  29. Stoichiometry Puzzle #3 of 3 Limiting reactant is a gas: What volume of butane gas is needed at room temperature (23oC) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825oC and a pressure of 1.04 atm? The reaction is: 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

  30. What volume of butane gas is needed at room temperature (23oC) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825oC and a pressure of 1.04 atm? The reaction is: 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) Asked:Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2

  31. Asked: Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2 Solve: C4H10 CO2

  32. Asked: Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2 Solve: C4H10 CO2

  33. Asked: Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2 Solve: C4H10 CO2 Use the mole ratio to substitute for nCO2

  34. Asked: Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2 Solve:

  35. Asked: Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2 Solve:

  36. Asked: Volume of C4H10 needed to produce 85.0 L of CO2 Given: C4H10: P = 0.984 atm, T = 23oC CO2: P = 1.04 atm, T = 825oC, V = 85.0 L Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2 Solve:

  37. Ideal gas law We have now solved stoichiometry problems involving gases and solids solutions other gases R = universal gas law

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