Gases Chapter 5 Lesson 2

Gas Stoichiometry Molar volume Things to remember volume M One mole of an ideal gas occupies 22.4L of volume at STP Things to remember Density = mass volume n = grams of substance = m molar mass M PV = mRT M

Gas Stoichiometry Example: A sample containing 15.0g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following equation: CO2(s)  CO2(g) How big will the balloon be (i.e. what is the volume of the balloon) at 22°C and 1.04 atm, after all of the dry ice has sublimed?

Gas Stoichiometry Moles of CO2(s): 15g 1 mol = 0.341 mol CO2(s) 44 g 0.341 mol CO2(s) 1 CO2(g) = 0.341 mol CO2(g) 1 CO2(s) = n

Gas Stoichiometry P = 1.04 atm V = ? n = 0.341 mol R = 0.0821 L•atm/mol• K T = 22C + 273 = 295K V = 7.94L

Gas Stoichiometry Limiting Reactant Example: 0.500L of H2(g) are reacted with 0.600L of O2(g) at STP according to the equation 2H2(g) + O2(g)  2H2O(g) What volume will the H2O occupy at 1.00 atm and 350°C?

Gas Stoichiometry H2: 0.5L 1 mol 2 H2O = 0.0223 mol H2O 22.4L 2 H2 O2: 0.6L 1 mol 2 H2O = 0.0536 mol H2O 22.4L 1 O2

Gas Stoichiometry H2: (1 atm)(V) = (0.0223mol)(0.0821 L•atm/mol• K)(623K) V = 1.14L O2: (1atm)(V) = (0.0536mol)(0.0821 L•atm/mol• K)(623K) V = 2.74L Limiting reactant and how much volume is produced!

Gas Stoichiometry Density/Molar Mass example: A gas at 34°C and 1.75 atm has a density of 3.40g/L. Calculate the molar mass of the gas. D = mass 3.40 = 3.40g V 1L P = 1.75 atm R = 0.0821 L•atm/mol• K V = 1L T = 34 + 273 = 307K m = 3.40 g M = x

Gas Stoichiometry (1.75atm)(1L) = (3.4g)(0.0821 L•atm/mol• K)(307K) M = molar mass M = 48.97 g/mol

Dalton’s Law of Partial Pressures (John Dalton, 1803) Statement of law “for a mixture of gases in a container, the total pressure exerted is the sum of the pressures each gas would exert if it were alone” It is the total number of moles of particles that is important, not the identity or composition of the gas particles.

Dalton’s Law of Partial Pressures Derivation Ptotal = P1 + P2 + P3 … P1 = n1RT P2 = n2RT P3 = n3RT …. V V V Ptotal = n1RT + n2RT + n3RT V V V Ptotal = (n1+n2+n3+…) (RT) V Ptotal = ntotal (RT)

Dalton’s Law of Partial Pressures Example: Oxygen gas is collected over water at 28C. The total pressure of the sample is 5.5 atm. At 28C, the vapor pressure of water is 1.2 atm. What pressure is the oxygen gas exerting? Ptotal = PO2 + PH2O 5.5 = x + 1.2 X = 4.3 atm

Dalton’s Law of Partial Pressures Mole Fraction The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture For an ideal gas, the mole fraction (x): x1 = n1 = P1 ntotal Ptotal

Dalton’s Law of Partial Pressures Example: The vapor pressure of water in air at 28°C is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28°C and 1.03 atm. XH2O = PH2O = 28.3 torr Pair 783 torr = 0.036

Dalton’s Law of Partial Pressures Example: A mixture of gases contains 1.5 moles of oxygen, 7.5 moles of nitrogen and 0.5 moles of carbon dioxide. If the total pressure exerted is 800 mmHg, what are the partial pressures of each gas in the mixture?

Dalton’s Law of Partial Pressures Calculate the total number of moles 1.5 + 7.5 + 0.5 = 9.5 moles Use mole fractions for each individual gas PO2: 800mmHg x 1.5 mol = 126 mmHg 9.5 mol PN2: 800mmHg x 7.5 mol = 632 mmHg PCO2: 800mmHg x 0.5 mol = 42 mmHg

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