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Stoichiometry balancing

Isothermal reactor. This example concerns an elementary, exothermic, second-order reversible liquid-phase reaction in a tubular reactor with a parabolic velocity distribution. Only the mole, rate law, and stoichiometry balance in the tubular reactor are required in ihi.s FEMLAB chemical engineering module. [Pg.1031]

Stoichiometry (Balancing Methods) Applied in Bioprocess Design... [Pg.382]

The balance of these two effects was found to depend delicately on the stoichiometry, pressure, and temperature. The results were used to develop a more comprehensive C0/H20/02/N0 reaction mechanism, incorporating the explicit fall-off behaviour of recombination reactions [46, 47],... [Pg.2118]

The reaction center has either to be spedfied when inputting a reaction into a database, or it has to be determined automatically. Specification on input is time-consuming but it can benefit from the insight of the human expert, particularly so if the reaction input is done by the primary investigator as is the case in an electronic notebook. Automatic determination of reaction centers is difficult, particularly so when incomplete readion equations are given where the stoichiometry of a reaction is not balanced see Section 3.1). One approach is to try first to complete the stoichiometry of a reaction equation by filling in the missing molecules such as water, N2, etc. and then to start with reaction center determination. A few systems for automatic reaction center specification are available. However, little has been published on this matter and therefore it is not discussed in any detail here. [Pg.175]

Stoichiometric relationships and calculations are important in many quantitative analyses. The stoichiometry between the reactants and products of a chemical reaction is given by the coefficients of a balanced chemical reaction. When it is inconvenient to balance reactions, conservation principles can be used to establish the stoichiometric relationships. [Pg.33]

Octabromodiphenyl Oxide. Octabromodiphenyl oxide [32536-52-0] (OBDPO) is prepared by bromination of diphenyl oxide. The degree of bromination is controlled either through stoichiometry (34) or through control of the reaction kinetics (35). The melting poiat and the composition of the commercial products vary somewhat. OBDPO is used primarily ia ABS resias where it offers a good balance of physical properties. Poor uv stabiUty is the primary drawback and use ia ABS is being supplanted by other brominated flame retardants, primarily TBBPA. [Pg.468]

For multiple reactions, material balances are required for each stoichiometry. [Pg.697]

An important engineering tool needed to implement Phase II is stoichiometry. Stoichiometry refers to the methods of material balances, which are defmed as precise accounts of the inputs and outputs of an operation. [Pg.364]

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. [Pg.276]

As discussed earlier, the admissible compositions may be selected as the lean-phase composition at some particular instant of time, or any other situation which is compatible with stoichiometry and mass-balance bounds such as Eqs. (8.20) and (8.21). Let us aihitrarily select the admissible composition of to be zero. [Pg.197]

For symbolic convenience we make use of the reaction variable x, which is the decrease in concentration of reactant A in time t. Because of the reaction stoichiometry, X is also the decrease in B concentration. The mass balance expressions are... [Pg.21]

Consider the balanced equation for glnconeogenesis in Section 23.1. Account for each of the components of this equation and the indicated stoichiometry. [Pg.772]

Stoichiometry in Reactive Systems. The use of molar units is preferred in chemical process calculations since the stoichiometry of a chemical reaction is always interpreted in terms of the number of molecules or number of moles. A stoichiometric equation is a balanced representation that indicates the relative proportions in which the reactants and products partake in a given reaction. For example, the following stoichiometric equation represents the combustion of propane in oxygen ... [Pg.334]

Chemistry is a quantitative science. This means that a chemist wishes to know more than the qualitative fact that a reaction occurs. He must answer questions beginning How much. . . The quantities may be expressed in grams, volumes, concentrations, percentage composition, or a host of other practical units. Ultimately, however, the understanding of chemistry requires that amounts be related quantitatively to balanced chemical reactions. The study of the quantitative relationships implied by a chemical reaction is called stoichiometry. [Pg.224]

The stoichiometry of growth and metabolism can also be described by elemental material balances. This approach can provide an insight into the potential of the organism for biomass or product production, and thus the scope for process improvement. [Pg.38]

An elemental material balance approach to growth stoichiometry requires an empirical formula for dry weight material ... [Pg.38]

Figure 9.9 shows the flow sheet for stoichiometry and material balance in a cell. According to the problem, two thirds of the carbon source goes to biomass. [Pg.247]

Rg. 9.9. Flow sheet for stoichiometry and material balance in a cell. [Pg.247]

Sometimes we need to know how much product to expect from a reaction, or how much reactant we need to make a desired amount of product. The quantitative aspect of chemical reactions is the part of chemistry called reaction stoichiometry. The key to reaction stoichiometry is the balanced chemical equation. Recall from Section H that a stoichiometric coefficient in a chemical equation tells us the relative amount (number of moles) of a substance that reacts or is produced. Thus, the stoichiometric coefficients in... [Pg.109]

When we do not know the changes, write one of them as x or a multiple of x and then use the stoichiometry of the balanced equation to express the other changes in terms of that x. [Pg.494]

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. [Pg.669]

The confusion was probably caused by the different stoichiometries of the balanced chemical equations, with one mole of sodium hydroxide reacting with one mole of dilute hydrochloric acid but two moles of the alkali reacting with one mole of dilute sulphuric acid. [Pg.164]

Most students demonstrated the abihty to translate from the sub-micro to the symbolic level by writing a balanced equation for the reaction in the question shown in Fig. 8.8. In determining the hmiting reagent, however, there were a lower number of correct responses than for the question in Fig. 8.9 based on stoichiometry. The difference in performance is even greater for part (c) of both questions... [Pg.183]

The following data were collected in an isothermal, constant-volume batch reactor. The stoichiometry is known and the material balance has been closed. The reactions are A B and A C. Assume they are elementary. Determine the rate constants kj and kn-... [Pg.252]

Sections 2- and 3- describe how to use the relationships among atoms, moles, and masses to answer how much questions about individual substances. Combining these ideas with the concept of a balanced chemical equation lets us answer how much questions about chemical reactions. The study of the amounts of materials consumed and produced in chemical reactions is called stoichiometry. [Pg.206]

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 ... [Pg.220]

All changes are related by stoichiometry. Each ratio of changes in amount equals the ratio of stoichiometric coefficients in the balanced equation. In the example above, the changes in amounts for H2 and N2 are in the ratio 3 1, the same as the ratio for the coefficients of H2 and N2 in the balanced equation. [Pg.220]

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 (. )... 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 (. )...
We could again apply the seven-step process in detail. Instead, we take a more compact approach. Begin by determining what species are present in the reaction mixture. Next, use the solubility guidelines to identify the precipitate. After writing the balanced net ionic reaction, use solution stoichiometry and a table of amounts to find the required quantities. [Pg.232]


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See also in sourсe #XX -- [ Pg.78 , Pg.79 ]

See also in sourсe #XX -- [ Pg.82 , Pg.83 , Pg.84 ]




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