Stoichiometry mole-volume conversions

Since the volume depends on conversion or time in a constant pressure batch reactor, consider the mole balance in relation to the fractional conversion X. From the stoichiometry. 

FIGURE 3.5 A flow diagram summarizing the use of molarity as a conversion factor between moles and volume in stoichiometry calculations. 

The concentration of a substance in solution is usually expressed as molarity (M), defined as the number of moles of a substance (the solute) dissolved per liter of solution. A solution s molarity acts as a conversion factor between solution volume and number of moles of solute, making it possible to carry out stoichiometry calculations on solutions. Often, chemicals are stored as concentrated aqueous solutions that are diluted before use. When carrying out a dilution, only the volume is changed by adding solvent the amount of solute is unchanged. A solution s exact concentration can often be determined by titration. 

These calculations show that the volume dependency of a gas-phase reaction is a function not only of the stoichiometry, but also of the inerts content of the reacting mixture. The sensitivity of volume to conversion is lowered as the inerts increase. The expansion factor, eA, is positive for reactions producing a net increase in moles, negative for a decrease in moles, and eA= 0 for reactions producing no net changes and at constant volume. 

Most chemical reactions that occur on the earth s surface, whether in living organisms or among inorganic substances, take place in aqueous solution. Chemical reactions carried out between substances in solution obey the requirements of stoichiometry discussed in Chapter 2, in the sense that the conservation laws embodied in balanced chemical equations are always in force. But here we must apply these requirements in a slightly different way. Instead of a conversion between masses and number of moles, using the molar mass as a conversion factor, the conversion is now between solution volumes and number of moles, with the concentration as the conversion factor. 

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. 

This is "simple" because the stoichiometry is one mole of reactant goes to one mole of product, and because the conversion of A to B follows first-order kinetics, as does the conversion of B back to A. Thus, when we assemble the two-component mass balance equations in a constant volume batch reactor, we find ... 

Consider reaction 4.3. No stoichiometric relationship is needed for calculating the conversion which is given simply by Equation 4.10. But the rate equation contains other concentrations as well, and these can be related to through the stoichiometry of the reaction. These relationships will depend on whether a volume change accompanies the reaction. Volume change can occur due to a change in number of moles + t a temperature, or pressure, or... 

Recall that the coefficients in a balanced equation give the relative number of moles of reactants and products, aexs (Section 3.6) To use this information, we must convert the masses of substances involved in a reaction into moles. When dealing with pure substances, as we did in Chapter 3, we use molar mass to convert between grams and moles of the substances. This conversion is not valid when working with a solution because both solute and solvent contribute to its mass. However, if we know the solute concentration, we can use molarity and volume to determine the number of moles (moles solute = M X F). T Figure 4.17 summarizes this approach to using stoichiometry for the reaction between a pure substance and a solution. 

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. 

When reactions occur in solution, reactant and product amounts are given in terms of concentration and volume. Molarity is the number of moles of solute dissolved in one liter of solution. Using molarity as a conversion factor, we apply the principles of stoichiometry to all aspects of reactions in solution. 

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. 

The two steps in the procedure can be combined so you can solve the problem in a single setup. The conversion factor between liters and moles is the molar volume at the temperature and pressure of the problem (see Section 14.5, particularly Example 14.9). From the ideal gas equation, molar volume is V/n, which, according to Equation 14.10, is RT/P. Thus, if you need to change moles to liters, multiply by V/n in the form RT/P. To change liters to moles, divide by RT/P, or multiply by its inverse, P/RT. This can be done in the same setup as the two steps in the stoichiometry path. The single setup for Example 14.14 becomes... 

We make the conversions between solution volumes and amounts of solute in moles using the molarities of the solutions. We make the conversions between amounts in moles of A and B using the stoichiometric coefficients from the balanced chemical equation. Example 4.8 demonstrates solution stoichiometry. 

The central conversion factor in Example 4-10 is the same as in previous stoichiometry problems—the appropriate stoichiometric factor. What differs from previous examples is that we use molarity as a conversion factor from solution volume to number of moles of reactant in a preliminary step preceding the stoichiometric factor. 

Such problems as this one involve many steps or conversions. Try to break the problem into simpler ones involving fewer steps or conversions. It may also help to remember that solving a stoichiometry problem involves three steps (1) converting to moles, (2) converting between moles, and (3) converting from moles. Use molarities and molar masses to carry out volume-mole conversions and gram-mole conversions, respectively, and stoichiometric factors to carry out mole-mole conversions. The stoichiometric factors are constructed from a balanced chemical equation. 

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