Gases volumes, stoichiometry problems

Be sure, especially in stoichiometry problems involving gases, that you are calculating the values such as volume and pressure of the correct gas. You can avoid this mistake by clearly labeling your quantities that means, mol of 02 instead of just mol. 

Another type of gas law problem involves stoichiometry. Gas stoichiometry problems are just like all other stoichiometry problems—you must use moles. In addition, one or more gas laws are necessary. Let s look at a gas stoichiometry problem. What volume, in liters of oxygen gas, collected over water, forms when 12.2 g ofKCl03 decompose according to the following equation ... 

In reaction stoichiometry problems involving gases, the ideal gas law provides a means to compute moles from pressure or volume data. 

This information will help you with a certain type of gas stoichiometry problem. When a gas reacts to produce another gas, you can use Gay-Lussac s law of combining volumes to find the volumes of the gases. The following Sample Problem shows you how. 

The best way to find out how to do a stoichiometry problem using the ideal gas law is to study an example. In the following Sample Problem, you will use a balanced equation and the ideal gas law to find the volume of a gas produced. (Refer to Chapter 4, section 4.1, if you want to review how to write balanced equations.)... 

We use the same strategy we used in the stoichiometry problems earlier in the book. That is, we compute the number of moles of CaC03 consumed and the number of moles of C02 produced. The moles of C02 can then be converted to volume by using the molar volume of an ideal gas. 

There are ways other than density to include volume in stoichiometry problems. For example, if a substance in the problem is a gas at standard temperature and pressure (STP), use the molar volume of a gas to change directly between volume of the gas and moles. The molar volume of a gas is 22.41 L/mol for any gas at STP. Also, if a substance in the problem is in aqueous solution, then use the concentration of the solution to convert the volume of the solution to the moles of the substance dissolved. This procedure is especially useful when you perform calculations involving the reaction between an acid and a base. Of course, even in these problems, the basic process remains the same change to moles, use the mole ratio, and change to the desired units. 

In gas stoichiometry problems, what is the bridge between amount in moles and volume ... 

The textbook s Web site has a section that describes a shortcut for working equation stoichiometry problems in which volume of one gas is converted into volume of another gas at the same temperature and pressure. 

Between this chapter and Chapter 10, we have now seen three different ways to convert between a measurable property and moles in equation stoichiometry problems. The different paths are summarized in Figure 13.10 in the sample study sheet on the next two pages. For pure liquids and solids, we can convert between mass and moles, using the molar mass as a conversion factor. For gases, we can convert between volume of gas and moles using the methods described above. For solutions, molarity provides a conversion factor that enables us to convert between moles of solute and volume of solution. Equation stoichiometry problems can contain any combination of two of these conversions, such as we see in Example 13.8. 

Convert between volume of gas and moles of gas in an equation stoichiometry problem using either the molar volume at STP, the ideal gas equation, or i as a conversion factor. 

In Chapter 3 we used relationships between amounts (in moles) and masses (in grams) of reactants and products to solve stoichiometry problems. When the reactants and/or products are gases, we can also use the relationships between amounts (moles, ri) and volume (V) to solve such problems (Figure 5.12). The following examples show how the gas laws are used in these calculations. 

In Chapters 3 and 4, we encountered many reactions that involved gases as reactants (e.g., combustion with O2) or as products (e.g., a metal displacing H2 from acid). From the balanced equation, we used stoichiometrically equivalent molar ratios to calculate the amounts (moles) of reactants and products and converted these quantities into masses, numbers of molecules, or solution volumes (see Figure 3.10). Figure 5.11 shows how you can expand your problem-solving repertoire by using the ideal gas law to convert between gas variables (F, T, and V) and amounts (moles) of gaseous reactants and products. In effect, you combine a gas law problem with a stoichiometry problem it is more realistic to measure the volume, pressure, and temperature of a gas than its mass. 

Tliis is a gas stoichiometry problem that requires knowledge of Avogadro s law to solve. Avogadro s law states that the volume of a gas is directly proportional to the number of moles of gas at constant temperatiue and pressure. 

This is a stoichiometry problem very much like the type we considered in Chapter 9. The oniy difference is that in this case, we want to calculate the volume of a gaseous product rather than the number of grams. To do so, we can use the relationship between moles and volume given by the ideal gas law. 

When we discussed quantitative aspects of chemical reactions in Chapter 4, we emphasized the importance of ratios of moles. The ideal gas law provides a relationship between the number of moles of a gas and some easily measurable properties pressure, volume, and temperature. So when gases are involved in a chemical reaction, the ideal gas law often provides the best way to determine the number of moles. Using the ideal gas law in a stoichiometry problem really doesn t involve any new ideas. It just combines two kinds of calculations that you ve already been doing. We ll still do the stoichiometric calculation in terms of mole ratios, as always, and we ll use the gas law to connect the number of moles of a gas with its temperature, pressure, and volume. 

We are asked to find the volume of a gas, and we are given its pressure and temperature. We ll assume that the gas behaves ideally. So if we knew the number of moles, we could easily use the gas law to get the volume we need. Looking a little closer, we should recognize this as a reaction stoichiometry problem because it asks us how much CO2 will be produced. The new wrinkle here is that it asks us to express the answer as a volume rather than as a mass or a number of moles. So we will first do a stoichiometric calculation to find the number of moles of CO2 produced and then use the gas law to find the volume of that amount of gas at the indicated temperature and pressure. As in any stoichiometry problem, we ll need to start with a balanced equation for the reaction to be sure we use the correct mole ratio. 

You should recognize this as a reaction stoichiometry problem because it is asking us how much CaC03 will be produced. As for any stoichiometry problem, we should write a balanced chemical equation for the reaction. Then convert the volume of gas into moles and proceed as usual. Because the gas volume given is at STP, we can use the molar volume we calculated above as a conversion factor. 

To carry out this problem requires first determining the amount of carbonate present (by multiplying the amount of ore by the fi action of carbonate). A stoichiometry problem can then be set up using the chemical equation to convert between moles of carbonate and moles of carbon dioxide. Finally, the volume can be determined using the ideal gas law, as long as the temperature and pressure are known or measurable. 

When working stoichiometry problems like the one in the preceding section involving the decomposition of potassium chlorate, the oxygen is normally collected over water by displacement and the volume is then measured. However, in order to get the pressure of just the oxygen, you have to subtract the pressure due to the water vapor. You have to mathematically dry out the gas. 

Stoichiometry Problems Involving Gas Volumes 53 Gas Mixtures Law of Partial Pressures... 

Solving stoichiometry problems involving gas volumes Given the volume (or mass) of one substance in a reaction,... 

To solve a stoichiometry problem that involves a gas whose volume is measured at STP, simply use 22.4 L/mol where you would have used molar mass if the amount had been given in grams. 

We now consider gas stoichiometry problems at temperatures and pressures other than STP. At STP you know that molar volume is 22.4 L/mol, and you have used it to solve those problems. In this section, you must first calculate the molar volume at the given temperature and pressure. Once you know the molar volume, you can solve all gas stoichiometry problems in exactly the same way. 

The stoichiometry path may be summarized as given quantity mol given mol wanted wanted quantity. In a gas stoichiometry problem, the first or third step in the path is a conversion between moles and liters of gas at a given temperature and pressure. If you are given volume, you must convert to moles if you find moles of wanted substance, you must convert to volume. These conversions are made with the ideal gas equation, PV = nRT. You have already made conversions like these. For example, in Example 14.3, you calculated the volume occupied by 0.393 mol N2 at 24°C and 0.971 atm. You used the ideal gas equation solved for V. 

In a gas stoichiometry problem, either the given quantity or the wanted quantity is a gas at specified temperature and pressure. The problem is usually solved in two steps, the order of which depends on whether the gas volume is the wanted quantity or the given quantity. 

This volume ratio is useful for stoichiometry problems when both the given and wanted quantities are gases measured at the same temperature and pressure. For example, consider the reaction 3 H2(g) + N2(g) 2 NH3(g). Let s calculate the volume of ammonia that will be produced by the reaction of 5 liters of N2, with both gases measured at STP. The equation coefficients, interpreted for gas volumes, tell us... 

Gas stoichiometry problems are usually solved in two steps. However, using RT/P for molar volume makes it possible to solve... 

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