The Chemical Equation and Stoichiometry

Your objectives in studying this section are to be able to  

Know the products of common reactions given the reactants. 

Calculate the stoichiometric quantities of reactants and products given the chemical equation. 

Define excess reactant, limiting reactant, conversion, degree of completion, and yield in a reaction. 

Identify the limiting and excess reactants and calculate the percent excess reactant(s), the percent conversion, the percent completion, and 

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A stoichiometry calculation is thus essentially a three-step procedure in which 1) the weight of D is divided by its formula weight to get moles of D, 2) the moles of D are converted to the moles of A by multiplying by the mole ratio a/d, as found in the chemical equation, and 3) the moles of A are converted to grams of A by multiplying by the formula weight of A. 

In Chapter 2, we described a chemical equation as a representation of what occurs when molecules react. We will now study chemical equations more closely to answer questions about the stoichiometry of reactions. Stoichiometry (pronounced stoy-key-om -e-tree ) is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most quantitative work in chemistry. In the next sections, we will use the industrial Haber process for the production of ammonia to illustrate stoichiometric calculations. 

The essential information implied by the chemical equation is the stoichiometry at the macroscopic level, ie, if a moles of M react, then b moles of B do also p moles of P are formed, etc. No inference should be made about behavior at the microscopic or atomic level, ie, there is no implication thatp molecules of P appear simultaneously. There may or may not be intermediates that appear and disappear in the course of the reaction. 

Strategy First (1), write the chemical equation for the reaction that occurs when strong acid or base is added. Then (2), apply the principles of stoichiometry to find the numbers of moles of weak acid and weak base remaining after reaction. Finally (3), apply the relation [H+] = X mhb/mb to find [H+] and then the pH. 

Step 3 Write the chemical equation for the neutralization reaction and use the reaction stoichiometry to find the amount of H. O ions (or OH ions if the analyte is a strong base) that remains in the analyte solution after all the added titrant reacts. Each mole of H30+ ions reacts with 1 mol OH ions therefore, subtract the number of moles of H30+ or OH ions that have reacted from the initial number of moles. 

To construct an overall rate law from a mechanism, write the rate law for each of the elementary reactions that have been proposed then combine them into an overall rate law. First, it is important to realize that the chemical equation for an elementary reaction is different from the balanced chemical equation for the overall reaction. The overall chemical equation gives the overall stoichiometry of the reaction, but tells us nothing about how the reaction occurs and so we must find the rate law experimentally. In contrast, an elementary step shows explicitly which particles and how many of each we propose come together in that step of the reaction. Because the elementary reaction shows how the reaction occurs, the rate of that step depends on the concentrations of those particles. Therefore, we can write the rate law for an elementary reaction (but not for the overall reaction) from its chemical equation, with each exponent in the rate law being the same as the number of particles of a given type participating in the reaction, as summarized in Table 13.3. 

Tables of amounts are useful in stoichiometry calculations for precipitation reactions. For example, a precipitate of Fe (OH) forms when 50.0 mL of 1.50 M NaOH is mixed with 35.0 mL of 1.00 M FeCl3 solution. We need a balanced chemical equation and amounts in moles to calculate how much precipitate forms. The balanced chemical equation is the net reaction for formation of Fe (OH)3 Fe (ag) + 3 OH (a g) Fe (OH)3 (. )...

The information for the calculation is organized around the chemical equation. Let x = mol H2 (or I2) that reacts. Then use stoichiometry to determine the amount of HI formed, in terms of x, and finally solve for x. 

C. Reactions and stoichiometry Provides the chemical reaction equations and stoichiometry. 

Attempts to define operationally the rate of reaction in terms of certain derivatives with respect to time (r) are generally unnecessarily restrictive, since they relate primarily to closed static systems, and some relate to reacting systems for which the stoichiometry must be explicitly known in the form of one chemical equation in each case. For example, a IUPAC Commission (Mils, 1988) recommends that a species-independent rate of reaction be defined by r = (l/v,V)(dn,/dO, where vt and nf are, respectively, the stoichiometric coefficient in the chemical equation corresponding to the reaction, and the number of moles of species i in volume V. However, for a flow system at steady-state, this definition is inappropriate, and a corresponding expression requires a particular application of the mass-balance equation (see Chapter 2). Similar points of view about rate have been expressed by Dixon (1970) and by Cassano (1980). 

Alternatively, the conservation of atomic species is commonly expressed in the form of chemical equations, corresponding to chemical reactions. We refer to the stoichiometric constraints expressed this way as chemical reaction stoichiometry. A simple system is represented by one chemical equation, and a complex system by a set of chemical equations. Determining the number and a proper set of chemical equations for a specified list of species (reactants and products) is the role of chemical reaction stoichiometry. 

This balanced equation can be read as 4 iron atoms react with 3 oxygen molecules to produce 2 iron(III) oxide units. However, the coefficients can stand not only for the number of atoms or molecules (microscopic level) but they can also stand for the number of moles of reactants or products. So the equation can also be read as 4 mol of iron react with 3 mol of oxygen to produce 2 mol ofiron(III) oxide. In addition, if we know the number of moles, the number of grams or molecules may be calculated. This is stoichiometry, the calculation of the amount (mass, moles, particles) of one substance in the chemical equation from another. The coefficients in the balanced chemical equation define the mathematical relationship between the reactants and products and allow the conversion from moles of one chemical species in the reaction to another. 

Unfortunately, OH and O concentrations in flames are determined by detailed chemical kinetics and cannot be accurately predicted from simple equilibrium at the local temperature and stoichiometry. This is particularly true when active soot oxidation is occurring and the local temperature is decreasing with flame residence time [59], As a consequence, most attempts to model soot oxidation in flames have by necessity used a relation based on oxidation by 02 and then applied a correction factor to augment the rate to approximate the effect of oxidation by radicals. The two most commonly applied rate equations for soot oxidation by 02 are those developed by Lee el al. [61] and Nagle and Strickland-Constable [62],... 

This equilibrium expression depends only on the stoichiometry of the reaction. By convention, chemists always write the concentrations of the products in the numerator and the concentrations of the reactants in the denominator. Each concentration term is raised to the power of the coefficient in the chemical equation. The terms are multiplied, never added. 

Question (b) is a matter of chemical kinetics and reduces to the need to know the rate equation and the rate constants (customarily designated k) for the various steps involved in the reaction mechanism. Note that the rate equation for a particular reaction is not necessarily obtainable by inspection of the stoichiometry of the reaction, unless the mechanism is a one-step process—and this is something that usually has to be determined by experiment. Chemical reaction time scales range from fractions of a nanosecond to millions of years or more. Thus, even if the answer to question (a) is that the reaction is expected to go to essential completion, the reaction may be so slow as to be totally impractical in engineering terms. A brief review of some basic principles of chemical kinetics is given in Section 2.5. 

Step 2 Calculate the moles of OH- ions in the volume of titrant added. Step 3 Write the chemical equation for the reaction between acid and base and use the reaction stoichiometry to calculate the number of moles of conjugate base (in this case, HC02- ions) formed by the reaction of the add with added base and the moles of weak acid (in this case, HCOOH) remaining. 

The order of a reaction cannot in general be predicted from the chemical equation a rate law is an empirical law. That is, a rate law is an experimentally determined characteristic of the reaction and cannot in general be written down from the stoichiometry of the chemical equation for the reaction. For instance, both the decomposition of N205 and that of N02 have a stoichiometric coefficient of 2 for the reactant, but one reaction is first order and the other is second order. The decomposition of ammonia also has a stoichiometric coefficient of 2 for the reactant, but its rate law is zero order. 

The chemical equation for an elementary reaction is a description of an individual molecular event that involves breaking and/or making chemical bonds. By contrast, the balanced equation for an overall reaction describes only the stoichiometry of the overall process, but provides no information about how the reaction occurs. The equation for the reaction of N02 with CO, for example, does not tell us that the reaction occurs by direct transfer of an oxygen atom from an N02 molecule to a CO molecule. 

At the beginning of Chapter 12, we raised three key questions about chemical reactions What happens How fast does it happen To what extent does it happen The answer to the first question is given by the stoichiometry of the balanced chemical equation, and the answer to the second question is given by the kinetics of the reaction. In this chapter, we ll look at the answer to the third question How far does a reaction proceed r... 

The sequence of conversions in Figure 18.20 is used to calculate the mass or volume of product produced by passing a known current through a cell for a fixed period of time. The key is to think of the electrons as a "reactant" in a balanced chemical equation and then to proceed as with any other stoichiometry problem. Worked Example 18.10 illustrates the calculations. Alternatively, we can calculate the current (or time) required to produce a given amount of product by working through the sequence in Figure 18.20 in the reverse direction, as shown in Worked Example 18.11. 

Stoichiometry is the series of calculations on the basis of formulas and chemical equations and will be covered in Chapter 4. The use of conversion factors is common even when the relative proportions are not fixed by a chemical formula. Consider a silver alloy used for jewelry production. (Alloys are mixtures of metals and, as mixtures, may be produced in differing ratios of the metals.) A particular alloy contains 86 percent silver. Factors based on this composition, such as... 

CO2 from combustion is found from the values calculated in part D and the stoichiometry of the chemical equations from part B. 

You have learned how to do stoichiometric calculations, using balanced chemical equations to find amounts of reactants and products. In these calculations, you assumed that the reactants and products occurred in the exact molar ratios shown by the chemical equation. In real life, however, reactants are often not present in these exact ratios. Similarly, the amount of product that is predicted by stoichiometry is not always produced. 

Recall that stoichiometry involves calculating the amounts of reactants and products in chemical reactions. If you know the atoms or ions in a formula or a reaction, you can use stoichiometry to determine the amounts of these atoms or ions that react. Solving stoichiometry problems in solution chemistry involves the same strategies you learned in Unit 2. Calculations involving solutions sometimes require a few additional steps, however. For example, if a precipitate forms, the net ionic equation may be easier to use than the chemical equation. Also, some problems may require you to calculate the amount of a reactant, given the volume and concentration of the solution. 

As was mentioned earlier in this section, heat is a reactant or product in most chemical reactions. It is possible for us to indicate the quantity of heat in the balanced equation and to treat it with the rules of stoichiometry that we already know. 

In general, the reaction order does not follow from the stoichiometry of the chemical equation. In the case of the catalytic decomposition of ammonia on a hot platinum wire, the reaction order is zero initially because the reaction occurs on the surface of the wire, and the surface coverage is independent of concentration. The rate of a zero-order reaction is independent of concentration until the reactant is nearly exhausted or until equilibrium is reached. 

The word stoichiometry is derived from the Greek stoicheion, which means first principle or element, and metron, which means measure. Stoichiometry describes the quantitative relationships among elements in compounds (composition stoichiometry) and among substances as they undergo chemical changes (reaction stoichiometry). In this chapter we are concerned with chemical formulas and composition stoichiometry. In Chapter 3 we shall discuss chemical equations and reaction stoichiometry. 

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