Download Gas Stoichiometry

Gas Stoichiometry
Balanced chemical equations can be used to relate
moles or grams of reactant(s) to products. When
gases are involved, these relations can be applied to
include volume.
Avogadro’s Law
• “Equal volumes of all gases, at the same temperature and
pressure, have the same number of molecules.”
Example of a Gas Stoichiometry Problem
Airbags in automobiles contain sodium azide (NaN3),
potassium nitrate, and silicon dioxide. (All are solids.)
1. Upon impact, the bag is inflated by the thermal
decomposition of sodium azide (NaN3) to sodium metal
and nitrogen gas.
2. Because sodium is toxic and very reactive, it reacts with the
potassium nitrate to produce potassium oxide and sodium
oxide, and (additional) nitrogen gas.
3. The metal oxides are removed by reacting with the silicon
dioxide to produce alkaline silicate (glass).
Question.
The driver’s airbag fills to 50-60 liters. Assuming the pressure
inside the airbag is 1 atm, calculate the number of grams of
each solid substance needed for a 50-L airbag.
Example 5.6
When acid is added to sodium bicarbonate (sodium hydrogen
carbonate), NaHCO3, the following reaction occurs:
NaHCO3(s) + H+(aq)  Na+(aq) + H2CO3(aq)
but H2CO3(aq) quickly decomposes to CO2 + H2O, so the
actual reaction is:
NaHCO3(s) + H+(aq)  Na+(aq) + CO2(g) + H2O(l)
All of the following experiments are performed with 2.45 M
HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
2.45 M HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
(a) If an excess of HCl is used, what volume of CO2 is produced?
NOTE:
The first solution that follows is ‘the long way around.’
After that, the solution is a shorter way that you should be able
to do. The strategies are the same.
Strategy
Balanced chemical equation (1 NaHCO3 = 1 CO2) (in pb.)
(i) Find amount of CO2 produced using stoichiometry
(ii) Static conditions so use PV = nRT
(note: you can’t use 22.4 L = 1 mol b/c it’s not at STP)
2.45 M HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
(a) If an excess of HCl is used, what volume of CO2 is produced?
NaHCO3(s) + H+(aq)  Na+(aq) + CO2(g) + H2O(l)
2.45 M HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
(a) If an excess of HCl is used, what volume of CO2 is produced?
NaHCO3(s) + H+(aq)  Na+(aq) + CO2(g) + H2O(l)
2.45 M HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
(b) If NaHCO3 is in excess, what volume of HCl is required to
produce 2.65 L of CO2?
2.45 M HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
(c) What volume of CO2 is produced when all of the NaHCO3 is
made to react with 50.0 mL HCl?
Strategy
(i) This involves determining the limiting reagent (reactant).
(ii) Convert nCO2
2.45 M HCl and 12.75 g of NaHCO3 at 732 mm Hg and 38oC.
(c) What volume of CO2 is produced when all of the NaHCO3 is made to react with 50.0
mL HCl?
NaHCO3(s) + H+(aq)  Na+(aq) + CO2(g) + H2O(l)
N.B.
Solving the previous problems, I repeated many steps that you
wouldn’t when solving the problem. For example, the number of
moles was calculated in (a) and used again in (c). On the AP
exam, you (and they) wouldn’t want you to repeat calculations.
Work smarter, not harder.