9.5 Gas Stoichiometry
Although all gases closely follow the ideal gas law PV = nRT under appropriate conditions, each gas is also a unique chemical substance consisting of molecular units that have definite masses. In this lesson we will see how these molecular masses affect the properties of gases that conform to the ideal gas law.
Following this, we will look at gases that contain more than one kind of molecule— in other words, mixtures of gases.
We begin with a review of molar volume and the E.V.E.N. principle, which is central to our understanding of gas mixtures.
In Reactions involving gaseous reactants and products we often report quantities in volumes at specific pressures and temperatures. We can convert this quantities to amounts ( in moles) using the ideal gas law. Then we can use the stoichiometric coefficients from the balanced equation to determine the stoichiometric amounts of other reactants and products.
The general form of these types of calculation s is
Volume A → amount of A(in mols) →amount of B( in moles) →molarity of B ( in desired units)
In case the reaction is accrued out at STP, we can use the molar volume at STP 22.4 L= 1mol. To convert between volume in liters and amount in moles.
Ideal Gas Law and Reaction Stoichiometry
Question:
In laboratory 25.0 g Barium peroxide, BaO2 was heated to produce gaseous oxygen. Oxygen gas was collected over water. Reaction produced an unknown gas that was collected at pressure 738 mm of Hg and 230C. Vapor pressure of water is 21 torr at 230C. How many Liters of oxygen was produced?
Molar mass of BaO2= 169.3 g/mol
2 BaO2 → 2BaO +2O2
Figure 9.57 gas Stoichiometry
Ref: commons.wikimedia.org/
Mols of BaO2= 25.0g/169.3 g/mol= 0.1477 mol
Mols of O2= 0.1477 mol * 2 mols of O2 = 0.1477 mol
2 mols of BaO2
Pressure of O2= 738-21= 717 torr= 717* 1 atm = 0.943 atm
760
Temperature= 23+273= 296 K
Applying ideal gas law: V= nRT/P= 0.1477mol *0.0821L-atm/mol-K*296K = 3.81 L
0.943 atm