1. CHAPTER 14
Gases
14.3 Stoichiometry and Gases

2. We can now solve stoichiometry problems involving gases
Ideal gas law
We can now solve stoichiometry problems involving gases
R = universal gas law

3. We can now solve stoichiometry problems involving gases
and
solids
solutions
other gases

4. Steps for solving stoichiometry problems

5. Steps for solving stoichiometry problems

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

8. 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:
2C20H O2 → 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:
2C20H O2 → 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 = 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 = 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 = g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
Solve:
125 g C20H42
0.442 moles 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 = 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 = 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 = 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 = 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 = 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 = g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
Solve:
8.85 moles CO2
1,170 L CO2

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

19. A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 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. Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 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 = 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. Asked: Pressure of H2 produced
Given: HCl: V = 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:
Volume
0.050 L HCl
1.25 M HCl

22. Asked: Pressure of H2 produced
Given: HCl: V = 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:
Volume
0.050 L HCl
1.25 M HCl
moles HCl

23. Asked: Pressure of H2 produced
Given: HCl: V = 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:
moles HCl

24. Asked: Pressure of H2 produced
Given: HCl: V = 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:
moles HCl
moles H2

25. Asked: Pressure of H2 produced
Given: HCl: V = 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:
moles H2

26. Asked: Pressure of H2 produced
Given: HCl: V = 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:
moles H2

27. Asked: Pressure of H2 produced
Given: HCl: V = 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:
moles H2
0.756 atm H2

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

29. 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
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)

30. 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
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 = atm, T = 23oC
CO2: P = 1.04 atm, T = 825oC, V = 85.0 L
Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2

31. C4H10 CO2 Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 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. C4H10 CO2 Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 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. Use the mole ratio to substitute for nCO2
Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 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 = 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 = 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 = 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. We have now solved stoichiometry problems involving gases
Ideal gas law
We have now solved stoichiometry problems involving gases
and
solids
solutions
other gases
R = universal gas law