Chem 161 Practice Problems 3

# Chem 161 Practice Problems 3 - N2(g 3H2(g 2NH3(g 20.0 mL mL...

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N 2 (g) + 3H 2 (g) → 2NH 3 (g) 20.0 mL ____ mL _____ mL Starting with 20.0 mL of N 2 , what volume of H 2 is needed for a complete reaction? What volume of NH 3 is formed? Assume all gases are at the same temperature and pressure. At the same temperature and pressure, equal volumes of gases contain equal amounts (moles). This means that volume ratios are equivalent to mole ratios for gases at the same temperature and pressure. 20 mL N 2 x 2 2 N 1 H 3 = 60 mL H 2 20 mL N 2 x 2 3 N 1 NH 2 = 40 mL NH 3 The fact that gases (at the same temperature and pressure) combine in simple whole-number volume ratios was first observed more than 200 years ago by Joseph Gay-Lussac, and is known as Gay-Lussac’s Law of Combining Volumes. 3O 2 (g) → 2O 3 (g) Starting with 12.2 L of O 2 , what volume of O 3 will be formed? Assume all gases are at the same temperature and pressure. 12.2 L O 2 x 2 3 O L 3 O L 2 = 8.1 L Note that the mole ratio of corresponds to the mole ratio as well as the volume ratio of gases (at the same temperature and pressure). Partial Pressure

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In a mixture of gases, each gas exerts its own independent pressure, called the partial pressure of the gas. The partial pressure is the pressure of a particular gas in a mixture, and would be the pressure exerted by that gas if all the other gases were removed. The partial pressure is proportional to the amount (moles) of each gas in the mixture. P A = V RT n A P B = V RT n B P C = V RT n C where P A , P B , P C , represent the partial pressures of gases A, B, C respectively. All gases in the mixture occupy the entire volume (V) of the container, and all are at the same temperature T. The ratio of two partial pressures is the same as the ratio of the moles of those gases. B A P P = B A n n The sum of the partial pressures is equal to the total pressure. P total = P A + P B + P C = V RT ) n n n ( C B A + + = V RT n total It is useful to compare the partial pressure of a given gas to the total pressure. total A P P = total A n n = x A (mole fraction of A) or: P A = x A P total A mixture of 16.0 g O 2 and 16.0 g CH 4 exerts a total pressure of 600 Torr. What is the partial pressure of each gas? 16.0 g O 2 x = 0.50 mol 16.0 g CH 4 x = 1.00 mol P CH 4 = x CH4 P total = = 400 Torr P O 2 = x O2 P total = = 200 Torr A container holds 1.00 mol CH 4 exerting a pressure of 0.60 atm. What amount of CO must be added to the container to increase the total pressure to 0.90 atm? Assume no CH escapes.
The extra CO 2 must exert 0.30 atm, (to increase the pressure from 0.60 atm to 0.90 atm) or half the 0.60 atm exerted by 1.00 mol CH 4 . Since pressure is proportional to amount, then 0.50 mol CO 2 must be added. Mathematically: 4 2 CH CO P P = 4 2 CH CO n n = x = 0.500 mol CO 2 Potassium chlorate when heated, decomposes into potassium chloride and oxygen. 2KClO

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Chem 161 Practice Problems 3 - N2(g 3H2(g 2NH3(g 20.0 mL mL...

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