Grade 10

Grade 10Stoichiometry and Chemical Calculations


Limiting and Excess Reactants


Chemistry often involves reactions where different substances (reactants) interact to form new substances (products). In many chemical reactions, you will have more of one reactant than is needed to completely react with another, leading to the concepts of limiting and excess reactants. Understanding these two concepts is fundamental to estimating the amount of product and identifying which reactant will be left after the reaction. This topic is especially important in stoichiometry, which is the study and calculation of the amounts of elements and compounds involved in chemical reactions.

What are the limiting and excess reactants?

In any chemical reaction, the limiting reactant is the substance that is completely consumed first. It decides how much product can be made because once it is used up, the reaction cannot continue. On the other hand, the excess reactant is the substance that is left over even after the reaction is complete. It is left over because there was not enough of the limiting reactant to be used up.

To understand these concepts more clearly, let's use a similar situation. Imagine that you are assembling toy cars, and each toy car requires 4 wheels and 1 body. If you have 10 bodies and 36 wheels, how many cars can you assemble?

Toy car assembly: Wheels required per car = 4 Bodies required per car = 1 Available: Bodies = 10 Wheels = 36

To find out how many cars you can assemble, determine which part (body or wheels) will finish first:

  • You can make 10 cars from 10 bodies, because you have exactly 10 bodies.
  • To make 10 cars you will need 4 wheels per car, meaning you will need a total of 40 wheels.
  • However, you only have 36 wheels.

This means that the number of cars you can build depends on the number of wheels you have. Wheels are a limited component, and bodies are additional.

In chemical reactions

To understand this concept in the context of chemistry, let's look at a real chemical reaction:

Example: The reaction of hydrogen gas with oxygen gas to form water

2H2 + O2 → 2H2O

According to the equation, 2 molecules of hydrogen (H2) react with 1 molecule of oxygen (O2) to form 2 molecules of water (H2O). If you have 4 moles of hydrogen and 1 mole of oxygen, how many moles of water will be formed?

Steps to identify limiting and excess reactants

  1. Calculate the mole ratio from the balanced equation:
    Mole ratio: 2 H2 : 1 O2 : 2 H2O
  2. Determine the amount of product formed from each reactant.
    From H2: 4 moles H2 × (2 mol H2O / 2 mol H2) = 4 moles H2O From O2: 1 mole O2 × (2 mol H2O / 1 mol O2) = 2 moles H2O
  3. The limiting reactant is the one that forms the least amount of product. Here it is O2, which forms only 2 moles of H2O.
  4. The excess reactant is H2.

Therefore, only 2 moles of water are produced, and the hydrogen remains unreacted.

Visual representation

To further illustrate limiting and excess reactants in reactions, consider this diagram:

+ 1 O2 (limited) → 2 H2O

Why is it important to identify the limiting reactant?

Determining the limiting reactant is important for several reasons:

  • Predicting product yield: By knowing the limiting reactant, you can accurately calculate how much product will be obtained from the reaction. This helps predict results and efficiency in experiments and industrial processes.
  • Efficient use of resources: Identifying the limiting reactant helps chemists use resources effectively and avoid wastage.
  • Cost-effectiveness in industry: In industrial chemical reactions, understanding which reactants limit production can help reduce costs by optimizing the amount of reactants used.

Examples and practice problems

Example: The reaction between nitrogen and hydrogen produces ammonia

N2 + 3H2 → 2NH3

Suppose you have 4 moles of N2 and 9 moles of H2, determine the limiting reactant.

  1. Determine the mole ratio:
    N2 : 3H2 : 2NH3
  2. Calculate the potential yield:
    From N2: 4 moles N2 × (2 mol NH3 / 1 mol N2) = 8 moles NH3 From H2: 9 moles H2 × (2 mol NH3 / 3 mol H2) = 6 moles NH3
  3. Determine the limiting reactant: H2 is the limiting reactant because it produces fewer moles of NH3.

Conclusion: 6 moles of NH3 can be produced from the excess of nitrogen.

Practice problem 1

Consider the reaction of carbon monoxide with hydrogen to form methanol:

CO + 2H2 → CH3OH

If you have 5 moles of CO and 8 moles of H2, find the limiting reactant and the amount of methanol produced.

Practice problem 2

In the combustion of propane (C3H8), given 10 moles of C3H8 and 20 moles of O2, find the limiting reactant and the excess reactant:

C3H8 + 5O2 → 3CO2 + 4H2O

Try solving these problems to solidify your understanding of limiting and excess reactants!

Conclusion

Understanding limiting and excess reactants is essential in the field of chemistry, particularly stoichiometry, in order to predict product formation and optimize chemical reactions. By identifying which reactants restrict the formation of products, scientists and industry can effectively manage resources, increase reaction efficiency, and reduce costs. Through practice, these concepts become easier to understand, paving the way for more advanced studies in chemistry.


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