Limiting Reactants The Problem

Now multiply by the percent-yield conversion factor that changes theoretical Sodium sulfate is used in dyeing yield to actual yield. textiles and in manufacturing glass 

The same reasoning can be applied to a chemistry question. Carbon reacts with oxygen to form carbon dioxide C(g) + 02(g) C02(g). Suppose you put three 

Go to http //now.brookscole.com/ cracolice3e and click Coached Problems for a tutorial on Percent Yield. 

The reactants combine in a one-to-one ratio. If you start with three atoms of carbon and two molecules of oxygen, the reaction will stop when the two oxygen molecules are used up. Oxygen, the reactant that is completely used up by the reaction, is called 

Since we cannot count individual particles, we again must turn to our macroscopic-particulate link, the mole. The amount of product is limited by the number of moles of the limiting reactant, and it must be calculated from that number of moles. If we start with three moles of carbon and two moles of oxygen, we have the same ratio of particles as when we start with three carbon atoms and two oxygen molecules. If two moles of the limiting reactant—oxygen—react, two moles of carbon dioxide are produced. According to the equation, each mole of carbon dioxide produced requires one mole of carbon to react. 

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Determine the limiting reactant The problem states that oxygen is in excess. To determine the limiting reactant, the number of moles of Na2C03, requiring molarity as a conversion factor, must be compared with the number of moles of NO, requiring a conversion factor based on the molar mass of NO. 

We will need the moles of HN02 in all the remaining steps in this problem. The moles of NaOH will be changing as we add more. The coefficients in the reaction are all ones, thus, we can simply compare the moles to find the limiting reactant. The sodium hydroxide, with the smaller number of moles is limiting. We can add the mole information to the balanced chemical equation. (The water, being neutral, will not be tracked.)... 

Complex stoichiometry problems should be worked slowly and carefully, one step at a time. When solving a problem that deals with limiting reactants, the idea is to find how many moles of all reactants are actually present and then compare the mole ratios of those actual amounts to the mole ratios required by the balanced equation. That comparison will identify the reactant there is too much of (the excess reactant) and the reactant there is too little of (the limiting reactant). 

Mujtaba and Macchietto (1997) have considered a maximum conversion problem for BREAD, subject to given product purity constraints. The reflux ratio is selected as the control parameters to be optimised for a fixed batch time so as to maximise the conversion of the limiting reactant. The optimal product amount, condenser and reboiler duties are also calculated. Referring to Figure 4.5 for CBD column the optimisation problem can be stated as ... 

Plan This is a limiting-reactant problem because the amounts of two reactants are given. After balancing the equation, we must determine the limiting reactant. The molarity (0.010 M) and volume (0.050 L) of the mercury(II) nitrate solution tell us the moles of one reactant, and the molarity (0.10 M) and volume (0.020 L) of the sodium sulfide solution tell us the moles of the other. Then, as in Sample Problem 3.10, we use the molar ratio to find the moles of HgS that form from each reactant, assuming the other reactant is present in excess. The limiting reactant is the one that forms fewer moles of HgS, which we convert to mass using the HgS molar mass. The roadmap shows the process. 

We can summarize the limiting-reactant problem as follows. Suppose you are given the amounts of reactants added to a vessel, and you wish to calculate the amount of product obtained when the reaction is complete. Unless you know that the reactants have been added in the molar proportions given by the chemical equation, the problem is twofold (1) you must first identify the limiting reactant (2) you then calculate the amount of product from the amount of limiting reactant. The next examples illustrate the steps. 

In problems involving limiting reactants, the first step is to detcamine which is the limiting reactant. After the limiting reactant has been identified, the rest of the problem can be solved using the approach outlined in Section 3.6. Consider the formation of methanol (CH3OH) firom carbon monoxide and hydrogen ... 

Determine the limiting reactant. This problem requires the addition of just enough H ions to react exactly with the OH ions present. Thus we need not be concerned with determiiung a limiting reactant. 

In section 11.3 vie showed that the difficult problem of solving the flux relations can be circumvented rather simply when the stoichiometric relations are satisfied by the flux vectors, but the treatment given there was limited to the case of a single Independent chemical reaction, when the stoichiometric relations permit all the flux vectors to be expressed in terms of any one of them. The question then arises whether any comparable simplification is possible v en the reactants participate in more than one independent reaction. 

Two product barrier layers are formed and the continuation of reaction requires that A is transported across CB and C across AD, assuming that the (usually smaller) cations are the mobile species. The interface reactions involved and the mechanisms of ion migration are similar to those already described for other systems. (It is also possible that solid solutions will be formed.) As Welch [111] has pointed out, reaction between solids, however complex they may be, can (usually) be resolved into a series of interactions between two phases. In complicated processes an increased number of phases, interfaces, and migrant entities must be characterized and this requires an appropriate increase in the number of variables measured, with all the attendant difficulties and limitations. However, the careful selection of components of the reactant mixture (e.g. the use of a common ion) or the imaginative design of reactant disposition can sometimes result in a significant simplification of the problems of interpretation, as is seen in some of the examples cited below. 

Section 1.5 described one basic problem of scaling batch reactors namely, it is impossible to maintain a constant mixing time if the scaleup ratio is large. However, this is a problem for fed-batch reactors and does not pose a limitation if the reactants are premixed. A single-phase, isothermal (or adiabatic) reaction in batch can be scaled indefinitely if the reactants are premixed and preheated before being charged. The restriction to single-phase systems avoids mass... 

To solve a quantitative limiting reactant problem, we identify the limiting reactant by working with amounts in moles and the stoichiometric coefficients from the balanced equation. For the ammonia synthesis, if we start with 84.0 g of molecular nitrogen and 24.2 g of molecular hydrogen, what mass of ammonia can be prepared First, convert from... 

A table of amounts is a convenient way to organize the data and summarize the calculations of a stoichiometry problem. Such a table helps to identify the limiting reactant, shows how much product will form during the reaction, and indicates how much of the excess reactant will be left over. A table of amounts has the balanced chemical equation at the top. The table has one column for each substance involved in the reaction and three rows listing amounts. The first row lists the starting amounts for all the substances. The second row shows the changes that occur during the reaction, and the last row lists the amounts present at the end of the reaction. Here is a table of amounts for the ammonia example ... 

The problem asks for a yield, so we identify this as a yield problem. In addition, we recognize this as a limiting reactant situation because we are given the masses of both starting materials. First, identify the limiting reactant by working with moles and stoichiometric coefficients then carry out standard stoichiometry calculations to determine the theoretical amount that could form. A table of amounts helps organize these calculations. Calculate the percent yield from the theoretical amount and the actual amount formed. 

Mixing the two solutions will produce 2.50 X 10 mol of Fe (0H)3 precipitate, which is 2.67 g. The mixed solution contains Na cations and Cl anions, too, but we can ignore these spectator ions in our calculations. Notice that this precipitation reaction is treated just like other limiting reactant problems. Examples and further illustrate the application of general stoichiometric principles to precipitation reactions. 

The problem asks for a percent yield, so we need to compare the actual yield and the theoretical yield. Information is provided about amounts of both starting materials, so this is a limiting reactant situation. 

Any of the types of problems discussed in Chapters 3 and 4 can involve gases. The strategy for doing stoichiometric calculations is the same whether the species involved are solids, liquids, or gases. In this chapter, we add the ideal gas equation to our equations for converting measured quantities into moles. Example is a limiting reactant problem that involves a gas. 

We have data for the amounts of both starting materials, so this is a limiting reactant problem. Given the chemical equation, the first step in a limiting reactant problem is to determine the number of moles of each starting material present at the beginning of the reaction. Next compute ratios of moles to coefficients to identify the limiting reactant. After that, a table of amounts summarizes the stoichiometry. 

Data are given for all reactants, so this is a limiting reactant problem. We must balance the chemical equation and then work with a table of molar amounts. 

C05-0040. Summarize the strategy for calculating final pressures of gases in a limiting reactant problem. 

The quantitative treatment of a reaction equilibrium usually involves one of two things. Either the equilibrium constant must be computed from a knowledge of concentrations, or equilibrium concentrations must be determined from a knowledge of initial conditions and Kgq. In this section, we describe the basic reasoning and techniques needed to solve equilibrium problems. Stoichiometry plays a major role in equilibrium calculations, so you may want to review the techniques described in Chapter 4, particularly Section 4- on limiting reactants. 

Again, we use the standard approach to an equilibrium calculation. In this case the reaction is a precipitation, for which the equilibrium constant is quite large. Thus, taking the reaction to completion by applying limiting reactant stoichiometry is likely to be the appropriate approach to solving the problem. 

In a similar way, electrochemistry may provide an atomic level control over the deposit, using electric potential (rather than temperature) to restrict deposition of elements. A surface electrochemical reaction limited in this manner is merely underpotential deposition (UPD see Sect. 4.3 for a detailed discussion). In ECALE, thin films of chemical compounds are formed, an atomic layer at a time, by using UPD, in a cycle thus, the formation of a binary compound involves the oxidative UPD of one element and the reductive UPD of another. The potential for the former should be negative of that used for the latter in order for the deposit to remain stable while the other component elements are being deposited. Practically, this sequential deposition is implemented by using a dual bath system or a flow cell, so as to alternately expose an electrode surface to different electrolytes. When conditions are well defined, the electrolytic layers are prone to grow two dimensionally rather than three dimensionally. ECALE requires the definition of precise experimental conditions, such as potentials, reactants, concentration, pH, charge-time, which are strictly dependent on the particular compound one wants to form, and the substrate as well. The problems with this technique are that the electrode is required to be rinsed after each UPD deposition, which may result in loss of potential control, deposit reproducibility problems, and waste of time and solution. Automated deposition systems have been developed as an attempt to overcome these problems. 

The treatment of the two-phase SECM problem applicable to immiscible liquid-liquid systems, requires a consideration of mass transfer in both liquid phases, unless conditions are selected so that the phase that does not contain the tip (denoted as phase 2 throughout this chapter) can be assumed to be maintained at a constant composition. Many SECM experiments on liquid-liquid interfaces have therefore employed much higher concentrations of the reactant of interest in phase 2 compared to the phase containing the tip (phase 1), so that depletion and diffusional effects in phase 2 can be eliminated [18,47,48]. This has the advantage that simpler theoretical treatments can be used, but places obvious limitations on the range of conditions under which reactions can be studied. In this section we review SECM theory appropriate to liquid-liquid interfaces at the full level where there are no restrictions on either the concentrations or diffusion coefficients of the reactants in the two phases. Specific attention is given to SECM feedback [49] and SECMIT [9], which represent the most widely used modes of operation. The extension of the models described to other techniques, such as DPSC, is relatively straightforward. 

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