Moles, excess and limiting reagents

Moles, excess and limiting reagents

Lessons

In this lesson, we will learn:
  • The importance of identifying the limiting reagent in reactions.
  • To identify by calculation the limiting and excess reagents in a chemical reaction.
  • To calculate quantities in excess.

Notes:

  • In the past few lessons we have learned to calculate amounts of substance used in reactions for the solid, gas and aqueous phases.
    Moles / stoichiometry test questions usually involve being given one quantity of a reactant or product and:
    • Converting the quantity of it from one unit into another unit;
    • Converting this unit into moles;
    • Using the molar or stoichiometric ratio of the reaction to find the moles of another substance;
    • Converting the moles of that new substance into another quantity for your final answer.
    You might only be asked about one reactant in a reaction, but remember for a reaction to go, every reactant in the equation must be present.

  • For a chemical reaction to happen, all the reactants must be present and available to react. If even one reactant is not present, the reaction will not happen; everything must be there. In real chemical reactions, this means that a reaction will go until one of the reactants has run out. When this happens, the reaction stops.
    • The chemical that you have the least amount of, or that runs out first is called the limiting reagent because its running out limits the reaction from happening any longer.
    • Reagents that are not limiting reagents are excess reagents or are “in excess”. We call it this because when the limiting reagent runs out, there will still be some of this reagent left over – an ‘excess’ amount of it.
    This is important to chemists as they plan reactions because we are doing them to obtain the products. If we only have x moles of a reactant, we can only expect y moles of product.

  • To find out the limiting reagent, you need to find the amount of product that can be made, with respect to each reactant involved. The reactant that would produce the smallest amount of product is the limiting reagent.
    To find the mass of excess reagent, find the amount of the excess reagent that reacts based on the amount of limiting reagent. Then, subtract that from the total amount of excess reagent available.
    Chemists do chemical reactions because we want the valuable products of them. This is why we find limiting reagents in terms of the amount of products we can get. This is also why finding excess reagent quantities is important; this excess is going to do nothing in the reaction unless we have more limiting reagent.

  • Worked example: Find the limiting reagent and quantity of excess.

  • Consider the reaction:

    HCl + NaOH \, \, NaCl + H2O

    Two aqueous solutions, one of 0.5 M HCl and 0.8 M NaOH can be mixed together to produce NaCl. 750 mL of the HCl solution is available, while 625 mL of the NaOH solution is available.

    Identify the limiting reagent in this reactant, and the quantity of excess reagent in mL.

    The first step in this problem is to find the number of moles of both reagents. Both are required, and one will run out before the other, so we need to calculate how much of both we have. The reagent with less moles is the limiting reagent.

    MolHCl=750mLHClMol \, HCl = 750 \, mL \, HCl \, 1LHCl1000mLHCl0.5molHCl1LHCl\large \frac{1 \, L \, HCl} {1000 \, mL \, HCl} \, * \, \frac{0.5 \, mol \, HCl } {1 \, L \, HCl } = 0.375 molHClmol \, HCl


    MolNaOH=625mLNaOHMol \, NaOH = 625 \, mL \, NaOH \, 1LNaOH1000mLNaOH0.8molNaOH1LNaOH\large \frac{1 \, L \, NaOH} {1000 \, mL \, NaOH} \, * \, \frac{0.8 \, mol \, NaOH } {1 \, L \, NaOH } = 0.5 molNaOHmol \, NaOH

    From these calculations we can see that the NaOH is in excess, so HCl is the limiting reagent. How much of the NaOH is in excess?

    0.5molNaOH\, mol \, NaOH - 0.375 molHClmol \, HCl = 0.125 molNaOH\, mol \, NaOH \, excess reagent


    Now we know that 0.125 moles of NaOH is in excess. This will not react because it doesn’t have any HCl to react with. How much volume of our NaOH solution contains 0.125 moles of NaOH, which we don’t need to use in the reaction?

    0.125molNaOH\, mol \, NaOH \, * \, 1LNaOH0.8molNaOH1000mLNaOH1LNaOH\large \frac{1 \, L \, NaOH} {0.8 \, mol \, NaOH} \, * \, \frac{1000 \, mL \, NaOH } {1 \, L \, NaOH } = 156.25 mLNaOH\, mL \, NaOH

    So this reaction will make as much product as possible with the entire solution of HCl, and 156.25 mL of the NaOH solution can be spared as this is in excess. This assumes the reaction goes to completion.

    Knowing your limiting reagent and the amount you have is important because a limit of reagents available puts a limit on the amount of products you can make too! Calculating the limiting reagent and the quantity of it manages expectations of our product yield, which we will learn about next lesson in Percentage yield and atom economy.
  • Introduction
    a)
    Recap: Moles and chemical reactions.

    b)
    The limiting reagents.

    c)
    How to find limiting and excess reagents.

    d)
    Worked example: identify the limiting reagent.


  • 1.
    Find the limiting reagent in a chemical reaction with known quantities.
    Consider the reaction:

    2H2+_2 + O2_2 →2H2_2 O

    a)
    50 g of O2_2 gas and 50 g of H2_2 gas were reacted together.
    Which reagent is the limiting reagent?

    b)
    What mass of the other reagent is in excess?


  • 2.
    Find the limiting reagent in a chemical reaction with known quantities.
    Consider the reaction:

    2C6_6H14+_{14} + 19O2_2 →12CO2+_2 + 14H2_2O

    a)
    If 120 g of O2_2 and 150 g of C6_6H14_{14} are reacted together, what is the limiting reagent?

    b)
    How many grams of the excess reagent are present in excess?


  • 3.
    Find the limiting reagent in a chemical reaction with known quantities.
    500 g of Fe2_2O3_3 is reacted with 750 g of C in the reaction:

    2Fe2_2O3(s)+_{3\;(s)} + 3C(s)_{\;(s)} →4Fe(s)+_{\;(s)} + 3CO2(g)_{2\;(g)}

    a)
    What mass of Fe is produced?

    b)
    What is the limiting reagent in this reaction?

    c)
    How many extra grams of this reagent are in excess?


  • 4.
    Find the limiting reagent in a chemical reaction with known quantities.
    45 g of Ca3_3(PO4_4)2_2 is reacted with 36 g C and 85 g SiO2_2 according to the reaction:

    2Ca3_3(PO4_4)2+_2 + 6SiO2+_2 + 10C→P4+_4 + 6CaSiO3+_3 + 10CO

    a)
    What mass of P4_4 can be made from these quantities?

    b)
    What is the limiting reagent?

    c)
    Find the excess mass of both excess reagents.