Chapter 9, Section 3, pages 320- 327
Stoichiometry and Cars
Objective
To relate volume calculations in stoichiometry to the inflation of air bags in cars
To use the limiting reactants to explain why fuel-air ratios affect engine performance
To compare the efficiency of pollution-control mechanisms in cars using percentage
yield.
Input
Stoichiometry and Air Bags
How do air bags inflate when you crash?
When you crash the impact causes the ignitor to fire causing reactants to combine
generating gas. The reactants must be present in the correct proportions so that the bag
does not over- or under-inflate. Here is just one of the many reactions used,
2NaN3 --------------> 2Na + 3N2 (g)
Other reactants make this reaction safe (HONORS)
Air-Bag Stoichiometry
Assume that you need 65.1 L N2 to inflate an airbag to the proper size. What mass of
NaN3 must be used?
1. What do you know?
2NaN3 --------------> 2Na + 3N2 (g
Volume of N2 = 65.1 L N2
Density of N2 = 0.92 g/L
Molar mass N2 = 28.02g/mol
Mass of reactant = ? NaN3
Molar mass of NaN3 = 65.02 g/mol
From the balanced equation 2 mol NaN3 = 3 mol N2
2. Calculate
? g NaN3 = 65.1 L N2 x 0.92 g N2 x 1 mol N2 x 2 mol NaN3 x 65.02 NaN3 = 93g
1L N2
28.02 g N2
3 mol N2
1 mol NaN2
Practice Problems pg. 322 #1-4
Stoichiometry and Engine Efficiency
The efficiency of a car engine depends on the gas (isooctane) to oxygen ratio.
2C8H18 + 25 O2 ----------> 16CO2 + 18 H20
The reactants must be in the above mole ratios, 2:25 or 1:12.5 in order for the engine
to run smoothly. Any other ratio may cause the engine not to run. Too much gas
engine “floods”, too little gas will stall. Remember that the amount of O2 in the air is
only 21%.
Air-Fuel Ratio
A cylinder in a car draws 0.5 L of air. How many ml. of isooctane shouid be injected
into the cylinder to completely react with the oxygen present? (Density isooctane =
0.692 g/ml and oxygen = 1.33g/ml. Air is 21% oxygen by volume)
1. What do you know?
2C8H18 + 25 O2 ----------> 16CO2 + 18 H20
volume of air = 0.5 L
Percettage by volume = 21%
Molar mass O2 = 32g/mol. C8H18 = 114.26g/mol
2 Plan you r work—need to know ml. of iso so you need to know moles of oxygen
3. Calculate
? mol O2=0.5 L air x 21 L O2 x 1.33g O2 x 1 mol O2 = 0.00436 mol O2
100L air
1L O2
32g O2
? ml C8H18 =0.0046 x 2 mol C8H18 x 114.26g C8H18 x 1 ml C8H18 = 5.76 x 10 -2
25 mol )2
1 mol C8H18
0.692g C8H18
Practice Problems pg. 324, #1-3
Stoichiometry and Pollution Control
Read yourself. This will not be covered in class.
Conclusion
Review key concepts.
Homework
Handout to be given out in class