STOICHIOMETRY
The Limiting Reagent

The Limiting Reagent and the Extent of Reaction Concepts

A general way to incorporate stoichiometric coefficients into chemical reaction calculations

Chemical reactions generally have two features that make calculations involving the amounts of material consumed or produced somewhat more involved than in the simplest case of a reaction that (a) "goes to completion," and (b) starts with exactly the right amounts of all reactants so that no reactants are left at the end. It may be that even if the reactants are present at the start in exactly the right proportion as dictated by the net reaction stoichiometry, some reactants may remain when the reaction "stops" (i.e., when chemical equilibrium has been attained). More common, however, is the case in which one reactant is present in stoichiometric excess, and the reaction will be limited and reach equilibrium when one of the other reactants is consumed (either completely or as much as the reaction allows). Here, we introduce a general way to approach the question "which reactant is limiting?" This approach will be useful in other reaction calculation scenarios we will encounter in a very few weeks!

Let's start off in general. Suppose compound A reacts with compound B to produce product compounds C and D according to the balanced net reaction (where a, b, c, and d are stoichiometric coefficients):

(For example, in the reaction 2 H₂ + O₂—> 2 H₂O, A is H₂, a is 2, B is O₂, b is 1, C is H₂O, c is 2, and there is no D.) Furthermore, let's imagine that we start the reaction with only arbitrary initial amounts of reactants A and B and no products. We will symbolize amounts (measured in mole units) as n in general, and in particular, we will symbolize initial amounts by n⁰. We write

Notice that whenever a moles of A disappears, b moles of B must have disappeared, too, and c moles of C appeared along with d moles of D.

If we have exactly the right stichiometric amounts of A and B and the reaction goes to completion, so that at equilibrium no A is left, no B is left, and stoichiometric amounts of C and D are produced, we can write the following equation (using n to represent the amounts at the end):

This equation expresses the idea that when a moles of A disappear, b moles of B must disappear, too, and c moles of C plus d moles of D must appear. It is a specific case of a more general expression, valid no matter how much A, B, C, and D are present at the start of the reaction. This expression defines a quantity x called the extent of reaction variable (or sometimes the degree of advancement variable). It is defined through the expressions

Note that for reactants, the numerators are of the form initial amount – later amount (expressing the idea that reactants disappear as the reaction goes from left to right), but for the products, they are later amount – initial amount. For any reaction under any initial conditions, x starts off at zero. (Note as well that x has units of moles, as do n and n⁰. Stoichiometric coefficients have no units.)

We will return to more general uses of this expression later, but for now, note that it gives us a fast and general way to find which reactant might be limiting and to find out how much of any product is produced as well as how much of any reactant in excess is left behind. We need only follow these steps:

1. Calculate the ratio of the initial amounts of each reactant (in moles) to its corresponding stoichiometric coefficient in the net reaction. The compound for which this ratio is smallest will be limiting. Call this smallest ratio x_max, because it will represent the maximum extent of reaction:

2. Multiply x_max by c, d, etc. (i.e., by all the product stoichiometric coefficients in turn) to calculate the amounts of each product that can be formed:

3. Finally, multiply x_max by a, b, etc. (i.e., by all the reactant stoichiometric coefficients in turn) and subtract these amounts from the initial amounts of each reactant, in turn, to calculate the reactant amounts left over at the end:

These steps assume the reaction "goes to completion" in the sense that they assume the limiting reagent is completely consumed. We will see later on how to treat the more general case.

Finally, we close with the example worked in lecture: we react 4.00 mol Al(s) with 5.00 mol I₂(s) in the net reaction

Step 1: For aluminum, we have (4.00 mol)/2 = 2.00 mol. For iodine, we have (5.00 mol)/3 = 1.67 mol. Thus, x_max is 1.67 mol, the smaller amount, and iodine is limiting.

Step 2: The amount of product aluminum triiodide we can form is 2x_max = 3.33 mol.

Step 3: The amount of aluminum left over is 4.00 mol – 2x_max = 0.67 mol.