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4.4 Reaction yields  (Page 2/9)

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mol HCl produced = 3 mol H 2 × 2 mol HCl 1 mol H 2 = 6 mol HCl

Complete reaction of the provided chlorine would produce

mol HCl produced = 2 mol Cl 2 × 2 mol HCl 1 mol Cl 2 = 4 mol HCl

The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant ( [link] ).

The figure shows a space-filling molecular models reacting. There is a reaction arrow pointing to the right in the middle. To the left of the reaction arrow there are three molecules each consisting of two green spheres bonded together. There are also five molecules each consisting of two smaller, white spheres bonded together. Above these molecules is the label, “Before reaction,” and below these molecules is the label, “6 H subscript 2 and 4 C l subscript 2.” To the right of the reaction arrow, there are eight molecules each consisting of one green sphere bonded to a smaller white sphere. There are also two molecules each consisting of two white spheres bonded together. Above these molecules is the label, “After reaction,” and below these molecules is the label, “8 H C l and 2 H subscript 2.”
When H 2 and Cl 2 are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.

View this interactive simulation illustrating the concepts of limiting and excess reactants.

Identifying the limiting reactant

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

3Si ( s ) + 2 N 2 ( g ) ⟶ Si 3 N 4 ( s )

Which is the limiting reactant when 2.00 g of Si and 1.50 g of N 2 react?

Solution

Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

mol Si = 2.00 g Si × 1 mol Si 28.09 g Si = 0.0712 mol Si
mol N 2 = 1.50 g N 2 × 1 mol N 2 28.02 g N 2 = 0.0535 mol N 2

The provided Si:N 2 molar ratio is:

0.0712 mol Si 0.0535 mol N 2 = 1.33 mol Si 1 mol N 2

The stoichiometric Si:N 2 ratio is:

3 mol Si 2 mol N 2 = 1.5 mol Si 1 mol N 2

Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.

Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

mol Si 3 N 4 produced = 0.0712 mol Si × 1 mol Si 3 N 4 3 mol Si = 0.0237 mol Si 3 N 4

while the 0.0535 moles of nitrogen would produce

mol Si 3 N 4 produced = 0.0535 mol N 2 × 1 mol Si 3 N 4 2 mol N 2 = 0.0268 mol Si 3 N 4

Since silicon yields the lesser amount of product, it is the limiting reactant.

Check your learning

Which is the limiting reactant when 5.00 g of H 2 and 10.0 g of O 2 react and form water?

Answer:

O 2

Got questions? Get instant answers now!

Percent yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield    of the reaction. In practice, the amount of product obtained is called the actual yield    , and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield    :

percent yield = actual yield theoretical yield × 100 %

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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