A balanced equation represents a situation in which a reaction proceeds to completion and all of the reactants are converted into products. Under nonstoichiometric conditions, the reaction continues only until one of the reactants is entirely consumed, and at this point the reaction stops. Under these conditions, it is important to know which reactant will be consumed first and how much product can be produced.

When a butane lighter is lit, for example, only a small amount of butane is released from the lighter and combined with a much larger amount of oxygen present in air.

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Under these conditions, the oxygen is considered an excess reactant because more is available than is required for reaction with butane. The butane is the limiting reactant, which means that it controls the amount of products produced in the reaction. When a nonstoichiometric reaction is complete, the limiting reactant is completely consumed and some amount of the excess reactant remains unreacted (Interactive Figure 3.4.1).

Due:
1/1/20 11:59 PM
Score: 0
Reaction of Zn with HCl

Consider a situation where 12 mol of butane is combined with 120 mol of oxygen. Under stoichiometric conditions, butane and oxygen react in a 2:13 mole ratio. The amount of oxygen needed to react with 10 mol of butane is

Because more oxygen is available than is needed to react with all of the available butane (120 mol available versus 78 mol needed), oxygen is the excess reactant. The amount of butane that would be needed to react with 120 mol of O2 is

Less butane is available than is needed to react with all of the oxygen available (12 mol available versus 18 mol needed), so butane is the limiting reactant. The amount of products that can be produced in the reaction is determined by the amount of limiting reactant present. For water,

The limiting reactant concept can be illustrated in the laboratory. For example, consider the reaction of iron(III) chloride with sodium hydroxide.

FeCl3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaCl(aq)

If a student starts with 50.0 g of FeCl3 and adds NaOH in 1-gram increments, the mass of Fe(OH)3 produced in each experiment can be measured. Plotting mass of Fe(OH)3 as a function of mass of NaOH added results in the graph shown in Interactive Figure 3.4.2.

Due:
1/1/20 11:59 PM
Score: 0
Mass of Fe(OH)3 as a function of mass of NaOH

As expected, as more NaOH is added to the FeCl3, increasing amounts of Fe(OH)3 are produced. But when more than 37 g of NaOH is added, the mass of Fe(OH)3 no longer changes. When 37 g (0.925 mol) of NaOH is added to 50 g (0.308 mol) of FeCl3, the stoichiometric ratio is equal to 3:1 and all reactants are consumed completely. When more than 37 g of NaOH is added to 50 g FeCl3, there is excess NaOH available (FeCl3 is limiting) and no additional Fe(OH)3 is produced.

Example Problem: Identify Limiting Reactants (Mole Ratio Method)

Identify the limiting reactant in the reaction of hydrogen and oxygen to form water if 61.0 g of O2 and 8.40 g of H2 are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.

Show Solution
Video Solution

You are asked to identify the limiting reactant and mass (in grams) of the excess reactant remaining after the reaction is complete.

You are given the mass of the reactants.

Step 1. Write a balanced chemical equation.

2 H2 + O2 → 2 H2O

Step 2. Determine the limiting reactant by comparing the relative amounts of reactants available.
Calculate the amount (in mol) of one of the reactants needed and compare that value to the amount available.

More O2 is needed (2.08 mol) than is available (1.91 mol), so O2 is the limiting reactant.

Alternatively, the amounts of reactants available can be compared to the stoichiometric ratio in the balanced equation in order to determine the limiting reactant.

=
>
ratio of reactants (available) ratio of reactants (balanced equation)

Here, the mole ratio of H2 to O2 available is greater than the mole ratio from the balanced chemical equation. Thus, hydrogen is in excess and oxygen is the limiting reactant.

Step 3. Use the amount of limiting reactant (O2) available to calculate the amount of excess reactant (H2) needed for complete reaction.

Step 4. Calculate the mass of excess reactant that remains when all the limiting reactant is consumed.

8.40 g H2 available - 7.70 g H2 consumed = 0.70 g H2 remains

Is your answer reasonable? More O2 is needed than is available, so it is the limiting reactant. There is H2 remaining after the reaction is complete because it is the excess reactant in the reaction.

Due: 1/1/20 11:59 PM
Score: 0
Due: 1/1/20 11:59 PM
Score: 0

An alternative method of performing stoichiometry calculations that involve limiting reactants is to simply calculate the maximum amount of product that could be produced from each reactant. (This method works particularly well for reactions with more than two reactants.) In this method, the reactant that produces the least amount of product is the limiting reactant.

Example Problem: Identify Limiting Reactants (Maximum Product Method)

Consider the reaction of gold with nitric acid and hydrochloric acid.

Au + 3 HNO3 + 4 HCl → HAuCl4 + 3 NO2 + 3 H2O

Determine the limiting reactant in a mixture containing 125 g of each reactant and calculate the maximum mass (in grams) of HAuCl4 that can be produced in the reaction.

Show Solution
Video Solution

You are asked to calculate the maximum mass of a product that can be formed in a reaction.

You are given the balanced equation and the mass of each reactant available.

In this case there are three reactants, so it is most efficient to calculate the maximum amount of product (HAuCl4) that can be produced by each one. The reactant that produces the least amount of HAuCl4 is the limiting reactant.

Step 1. Write the balanced chemical equation.

Au + 3 HNO3 + 4 HCl → HAuCl4 + 3 NO2 + 3 H2O

Step 2. Use the molar mass of each reactant and the stoichiometric factors derived from the balanced equation to calculate the amount of HAuCl4 that can be produced by each reactant.

Gold is the limiting reagent. The maximum amount of HAuCl4 that can be produced is 216 g.

Due: 1/1/20 11:59 PM
Score: 0
Due: 1/1/20 11:59 PM
Score: 0
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