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Published byKeenan Plumlee Modified over 3 years ago

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Sometimes a gas is produced in a chemical reaction…. That gas is bubbled through water and collected in a separate tube or container The total pressure inside the container, though, is not just the gas Some water has evaporated and is adding pressure to the gas The total pressure inside the container, then, is the pressure of the gas and the pressure of the water vapor that has evaporated The top of the container contains the gas produced from the reaction and water vapor

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Oxygen can be prepared from heating potassium chlorate. How many moles of oxygen would be obtained from 1.300 grams of potassium chlorate? If the oxygen is collected over water at 23 0 C and a total pressure of 745 mm Hg, what volume would the oxygen occupy? We do the same thing we always do – except one small change in the pressure! Write a chemical reaction: 2 KClO 3 (s) -----------> 2 KCl (s) + 3 O 2 (g) Convert to moles: 1.300 g KClO 3 x 1 mole KClO 3 /122.55 g =.0106 moles KClO 3 Set up a ratio from the reaction:.0106 moles KClO 3 x 3 moles O 2 / 2 moles KClO 3 =.0159 moles O 2 PV/nT = R However, we need to adjust the pressure – it tells us in the problem that the total pressure inside the container is 745 mm Hg – this pressure is both the gas created and the water that has evaporated!

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KClO 3 (s) -----------> KCl (s) + 3 O 2 (g).0159 moles O 2 has to be converted into liters to solve the problem However, we are not at standard temperature and pressure We need to use the ideal gas law The pressure given in the problem - 745 mm Hg – is the pressure of the O 2 and the water that has evaporated and is adding to the pressure! We have to subtract out the water vapor pressure from the total pressure to get the pressure of just the gas!

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We need a table that tells us what the evaporated pressure of water is at 23 0 C! The table to the right tells us that at 23 0 C, the vapor pressure due to water evaporating is 21.0 mm Hg. We need to subtract this from the total pressure – this will tell us how much of the pressure is due to just the gas! P gas = 745 mm Hg – 21.0 mm Hg P gas = 724 mm Hg We use this pressure to finish the problem!

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KClO 3 (s) -----------> KCl (s) + 3 O 2 (g) PV/nT = R =.0821 L atm/ K mole Make sure you convert all units appropriately! P gas = 724 mm Hg x 1 atm/760 mm Hg =.953 atm T = 23 0 C + 273 = 296 K n =.0159 moles O 2 V = ? (.953 atm)(x)/(.0159 moles O 2 )(296 K) =.0821 L atm/ K mole V =.405L O 2

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Density of gases and Dumass Law Years ago, it was easier to calculate the density of a gas in order to help identify it, rather than figure out moles! Moles were not easy to calculate. Density = mass/volume If you look at the ideal gas law: PV/nT = R And rearrange: PV = nRT n, or moles, is equal to mass/molar mass (M m ) We know this from the mole map! Old lab – figuring out moles, hard. Figuring out density, easy.

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Density of gases and Dumass Law PV = nRT Substituting in mass/M m for n gives you: PV = mass x R x T M m Multiplying both sides by M m gives you: M m x P x V = mass x R x T Dividing both sides by V, you get: M m x P = mass x R x T V mass/V is density, so this can be simplified into: M m x P = D x R x T Jean Baptist Andre Dumas

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M m x P = D x R x T This is called Dumas Law! It is more useful sometimes than the traditional gas law because it has density in it, which is easy for a scientist to measure! Lets look at a homework problem! What is the density of elemental oxygen in grams per liter at 25 0 C and.850 atm? Using Dumas Law, we are solving for D Elemental oxygen, or O 2, has a molar mass of 32 g/mole T needs to be converted into Kelvin – T = 25 0 C + 273 = 298 K (32 g/mole)(.850 atm) = (D)(.0821 L atm/K mole)(298 K) Solving, we get D = 1.11 g/L The only units left are g/L, units for density!

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5.4 – 5.5: Applying The Ideal Gas Law

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