1. Ppt 18a, Continuation of Gases
Stoichiometry when gases are involved
Kinetic Molecular Theory (begin)
Ppt18a

2. Gas Stoichiometry
Remember, if it’s a stoichiometry problem involving a chemical reaction, you will nearly always need to use a mole ratio from the balanced chemical equation at some point in the problem!
To get to or from moles, use:
a) MM (gmol), b) M (mol L), & NOW……c) ideal gas equation, etc.
DON’T FORGET THE STOICHIOMETRY PART OF A STOICHIOMETRY PROBLEM!!
Don’t use PV = nRT unless it’s a GAS!!! 
Ppt18a

3. Gas Stoichiometry
How many grams of NaN3 (MM = 65.0 g/mol) are needed to fill a car air bag (assume V = 60.0 L) to 1.20 atm at 35C ?
2 NaN3(s)  2 Na(s) + 3 N2(g)
Ans g (if you got 185 then you forgot to do the stoichiometry part of the problem! The moles you get from n = PV/RT are moles of N2, not the moles of NaN3(s)!!)
See also “For Practice 5.12, p. 204, Tro”
Ppt18a

4. 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
Gas Stoichiometry
Note that if T and P are kept constant, then volume is proportional to moles.
- “Equal volumes of different gases at the same T and P have the same number of moles” (and if V = 2x, then twice as many moles
- This means that one can use the coefficient ratios (mole ratios) as VOLUME ratios (as a shortcut)
Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
Ppt18a

5. Kinetic Molecular Theory (KMT)
It’s a theory  Explains Laws (behavior)
Specifically, KMT is a model that explains (is consistent with and predicts) the physical properties of gases in terms of the motion of individual gas “particles”.
“Tiny superball” analogy / image
Ppt18a

6. Animations of KMT
--allows changes in mass / particle and gas mixtures
Ppt18a

7. Kinetic Molecular Theory—formal postulates
(Recall the “superball” analogy!):
Gas “particles” (atoms or molecules) move in straight lines until they collide with something; Collisions with a surface are the cause of the pressure exerted on it.
Particle volume is negligible (technically, zero) compared to gas volume (vessel volume)
Distance between particles is HUGE compared to particle diameter; Most volume is “empty space”
Gas collisions are perfectly elastic & particles do not exert any forces on one another between collisions
Average Kinetic Energyparticle  Kelvin Temperature
Ppt18a