Limiting and Excess Reactants

In industrial reactors you will rarely find exact stoichiometric amounts of materials used. To make a desired reaction take place or to use up a costly reactant, excess reactants are nearly always used. The excess material comes out together with, or perhaps separately from, the product and sometimes can be used again. The limiting reactant is defined as the species in a chemical reaction that

theoretically would be the first to be completely consumed if the reaction were to proceed to completion according to the chemical equation—even if the reaction does not proceed to

completion! All of the other reactants are called excess reactants. For example, using the chemical reaction equation in Example 5.2,

C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

if 1 g mol of C₇H₁₆ and 12 g mol of O₂ are mixed so as to react, C₇H₁₆ would be the limiting reactant even if the reaction does not take place. The amount of the excess reactant O₂ would be calculated as 12 g mol of initial reactant less the 11 g mole needed to react with 1 g mol of C₇H₁₆, or 1 g mol of O₂. Therefore, if the reaction were to go to completion, the amount of product that would be produced is controlled by the amount of the limiting reactant, namely, C₇H₁₆ in this example.

As a straightforward way of determining which species is the limiting reactant, you can calculate the maximum extent of reaction, a quantity that is based on assuming the complete reaction of each reactant. The reactant with the smallest maximum extent of reaction is the limiting reactant. For Example 5.2, for 1 g mol of C₇H₁₆ and 12 g mol of O₂, you can calculate

Example 5.5. Calculation of the Limiting and Excess Reactants Given the Mass of Reactants In this example let’s use the same data as in Example 5.4. The basis is the same and the figure is the same.

a. What is the maximum number of grams of CH₂O that can be produced? b. What is the limiting reactant?

c. What is the excess reactant?

Solution

The first step is to determine the limiting reactant by calculating the maximum extent of reaction based on the complete reaction of both CO₂ and H.

You can conclude that (b) H is the limiting reactant, and that (c) CO₂ is the excess reactant. The excess CO₂ is (0.041 – 0.014) = 0.027 g mol. To answer question a, the maximum amount of CH₂O that can be produced is based on assuming complete reaction of the limiting reactant:

Finally, you should check your answer by working from the answer to the given reactant, or,

alternatively, by summing up the mass of the C and the mass of excess H. What should the sums be? 5.2.3. Conversion and Degree of Completion

Conversion and degree of completion are terms not as precisely defined as are the extent of reaction and limiting and excess reactant. Rather than cite all of the possible usages of these terms, many of which conflict, we shall define them as follows: Conversion (or the degree of completion) is the fraction of the limiting reactant in the feed that is converted into products. Conversion is related to the degree of completion of a reaction. The numerator and denominator of the fraction contain the same units, so the fraction conversion is dimensionless. Thus, percent conversion is

For example, for the reaction equation used in Example 5.2, if 14.4 kg of CO₂ are formed in the reaction of 10 kg of C₇H₁₆, you can calculate the percent of the C₇H₁₆ that is converted to CO₂ (reacts) as follows:

The conversion can also be calculated by using the extent of reaction as follows: Conversion is equal to the extent of reaction based on the formation of CO₂ (i.e., the actual extent of reaction) divided by the extent of reaction, assuming complete reaction of C₇H₁₆ (i.e., the maximum possible extent of reaction):

5.2.4. Selectivity

Selectivity is the ratio of the moles of a particular (usually the desired) product produced to the moles of another (usually undesired or by-product) product produced in a single reaction or group of reactions. For example, methanol (CH₃OH) can be converted into ethylene (C₂H₄) or propylene (C₃H₆) by the reactions

2 CH₃OH → C₂H₄ + 2H₂O 3 CH₃OH → C₃H₆ + 3H₂O

Of course, for the process to be economical, the value of the products has to be greater than the value of the reactants. Examine the data in Figure 5.1 for the concentrations of the products of the reactions. What is the selectivity of C₂H₄ relative to the C₃H₆ at 80% conversion of the CH₃OH? Proceed

upward at 80% conversion to get for C₂H₄ ≅ 19 mol % and for C₃H₆ ≅ 8 mol %. Because the basis for both values is the same, you can compute the selectivity 19/8 ≅ 2.4 mol C₂H₄ per mol C₃H₆.

Figure 5.1. Products from the conversion of ethanol

5.2.5. Yield

No universally agreed-upon definitions exist for yield—in fact, quite the contrary. Here are three common ones:

• Yield (based on feed): The amount (mass or moles) of desired product obtained divided by the amount of the key (frequently the limiting) reactant fed.

• Yield (based on reactant consumed): The amount (mass or moles) of desired product obtained divided by the amount of the key (frequently the limiting) reactant consumed.

• Yield (based on 100% conversion): The amount (mass or moles) of a product obtained divided by the theoretical (expected) amount of the product that would be obtained based on the limiting reactant in the chemical reaction equation(s) if it were completely consumed. Note that this is a fractional (dimensionless) yield because the numerator and denominator have the same units, whereas the previous two definitions of yield are not dimensionless.

Why doesn’t the actual yield in a reaction equal the theoretical yield predicted from the chemical reaction equation? Several reasons exist:

• Impurities among the reactants • Leaks to the environment • Side reactions

• Reversible reactions

As an illustration, suppose you have a reaction sequence as follows:

with B being the desired product and C the undesired one. The yield of B according to the first two definitions is the moles (or mass) of B produced divided by the moles (or mass) of A fed or

consumed. The yield according to the third definition is the moles (or mass) of B actually produced divided by the maximum amount of B that could be produced in the reaction sequence (i.e., complete conversion of A to B). The selectivity of B is the moles of B divided by the moles of C produced. The terms yield and selectivity are terms that measure the degree to which a desired reaction

proceeds relative to competing alternative (undesirable) reactions. As a designer of equipment, you want to maximize production of the desired product and minimize production of the unwanted

products. Do you want high or low selectivity? Yield?

The next example shows you how to calculate all of the terms discussed in this section. Example 5.6. Calculation of Various Terms Pertaining to Reactions

Semenov [N. N. Semenov, Some Problems in Chemical Kinetics and Reactivity, Vol. II, Princeton University Press, Princeton (1959), pp. 39–42] described some of the chemistry of alkyl chlorides. The two reactions of interest for this example are

C₃H₆ is propene (MW = 42.08).

C₃H₅Cl is allyl chloride (3-chloropropene) (MW = 76.53).

C₃H₆Cl₂ is propylene chloride (1,2-dichloropropane) (MW = 112.99).

Table E5.6

Based on the product distribution in Table E5.6, assuming that the feed consisted only of Cl₂ and C₃H₆, calculate the following:

a. How much Cl₂ and C₃H₆ were fed to the reactor in gram moles? b. What was the limiting reactant?

c. What was the excess reactant?

d. What was the fraction conversion of C₃H₆ to C₃H₅Cl? e. What was the selectivity of C₃H₅Cl relative to C₃H₆ Cl₂?

f. What was the yield of C₃H₅Cl expressed in grams of C₃H₅Cl to the grams of C₃H₆ fed to the reactor?

g. What was the extent of reaction of Reactions (1) and (2)?

Solution Steps 1-4

Examination of the problem statement reveals that the amount of feed is not given, and consequently you have to calculate the gram moles fed to the reactor even if the amounts were not asked for. The molecular weights were given. Figure E5.6 illustrates the process as an open-flow system.

Figure E5.6 Step 5

A convenient basis is what is given in the product list in Table E5.6. Steps 7–9

Use the chemical equations to calculate the moles of species in the feed. Start with the Cl₂. Reaction (1):

a. Total Cl₂ fed

What about the amount of C₃H₆ in the feed? From the chemical equations you can see that if 29.1 g mol of Cl₂ reacts in total by Reactions (1) and (2), 29.1 g mol of C₃H₆ must react. Since 651.0 g mol of C₃H₆ exist unreacted in the product, 651.0 + 29.1 = 680.1 g mol of C₃H₆ were fed to the reactor.

You can check those answers by adding up the gram moles of Cl, C, and H in the product and comparing the value with that calculated in the feed:

In product: Cl 2(141.0) + 1(4.6) + 2(24.5) + 1(4.6) = 340.2 C 3(651) + 3(4.6) + 3(24.5) = 2040.3 H 6(651) + 5(4.6) + 6(24.5) + 1(4.6) = 4080.6 In feed: Cl 2(170.1) = 340.2 OK C 3(680.1) = 2040.3 OK H 6(680.1) = 4080.6 OK

We will not go through detailed analysis for the remaining calculations but simply determine the desired quantities based on the data prepared for parts a, b, and c. In this particular problem, since both reactions involve the same reaction stoichiometric coefficients, both reactions will have the same limiting and excess reactants:

Thus, C₃H₆ was the excess reactant and Cl₂ the limiting reactant. d. The fraction conversion of C₃H₆ to C₃H₅Cl was

e. The selectivity was

g. Because C₃H₅Cl is produced only by the first reaction, the extent of reaction of the first reaction is

Because C₃H₆Cl₂ is produced only by the second reaction, the extent of reaction of the second reaction is

Self-Assessment Test

Questions