Module 3 Intro
1. Module 3 Intro
1.17. Page 4
Module 3—Behaviour of Gases
Self-Check
SC 2. Complete the following table.
Word Equation |
hydrogen |
+ |
chlorine |
→ |
hydrogen chloride |
Balanced Equation |
H2(g) |
+ |
|
→ |
2 HCl(g) |
Ratio |
1 |
|
|
|
2 |
Volume |
|
|
1.1 L |
|
|
SC 3.
The Breitling Orbiter 3 uses the combustion of propane and oxygen to heat gases in the two chambers of the high-altitude balloon. Complete the following table to investigate the proportion by which propane and oxygen are used in a combustion reaction.
Word Equation |
propane |
+ |
oxygen |
→ |
carbon dioxide |
+ |
water vapour |
Balanced Equation |
|
+ |
5 O2(g) |
→ |
3 CO2(g) |
+ |
4 H2O(g) |
Ratio |
|
|
5 |
|
3 |
|
4 |
Volume |
|
|
|
|
|
|
80 mL |
SC 4. Use the law of combining volumes to find the volume of oxygen required for the complete combustion of 83.0 mL of propane. Assume the pressure and temperature are held constant.
SC 5. Butane gas is combusted to produce carbon dioxide and water vapour. If 2.4 mol of butane reacts, how many moles of water vapour will be produced?
SC 6. Complete question 5(a) on page 168 of your textbook.
Self-Check Answers
SC 2.
Word Equation |
hydrogen |
+ |
chlorine |
→ |
hydrogen chloride |
Balanced Equation |
H2(g) |
+ |
Cl2(g) |
→ |
2 HCl(g) |
Ratio |
1 |
|
1 |
|
2 |
Volume |
1.1 L |
|
1.1 L |
|
2.2 L |
SC 3.
Word Equation |
propane |
+ |
oxygen |
→ |
carbon dioxide |
+ |
water vapour |
Balanced Equation |
C3H8(g) |
+ |
5 O2(g) |
→ |
3 CO2(g) |
+ |
4 H2O(g) |
Ratio |
1 |
|
5 |
|
3 |
|
4 |
Volume |
20 mL |
|
100 mL |
|
60 mL |
|
80 mL |
SC 4. First, write the balanced chemical formula.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
The ratio of oxygen to propane is: 5 molOxygen : 1 mol propane or
5 mLOxygen : 1 mL propane if the gases are at the same temperature and pressure.
Therefore, 83 mLpropane ( 5 mLoxygen / 1 mLpropane) = 415 mLoxygen
SC 5. First, write the balanced chemical formula.
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
The ratio of water vapour to butane is: 10 molwater vapour : 2 mol butane
Therefore, 2.4 moLbutane ( 10 moLwater vapour / 2 moLbutane) = 12moLwater vapour
SC 6. Question 5 (a) on page 168 of your textbook.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
V=100 L V = ? V= ? V= ?
Voxygen = 100Lammonia (5 Loxygen / 4 Lammonia) = 125 Loxygen
Vnitrogen monoxide = 100Lammonia (4 Lnitrogen monoxide / 4 Lammonia) = 100 Lnitrogen monoxide
Vwater vapour = 100Lammonia (6 Lwater vapour / 4 Lammonia) = 150 Lwater vapour