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Limiting Reagents & Reactants in Excess

Key Concepts

  • The limiting reagent is the reactant that is completely used up during the chemical reaction.

  • The reactant that is in excess is the reactant that is not completely used up during the chemical reaction, that is, there is some of this reactant left over.

Deciding which reactants are the limiting reagents and the reactants in excess

  1. Write the balanced chemical equation for the chemical reaction

  2. Calculate the available moles of each reactant in the chemical reaction

  3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

  4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

  5. The limiting reagent is the reactant that will be completely used up during the chemical reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion.

Examples

Moles of reactants given

Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl

  1. Write the balanced chemical equation for the chemical reaction

        Zn + 2HCl → ZnCl2 + H2

  2. Calculate the available moles of each reactant in the chemical reaction
    moles of Zn = 0.5 moles of HCl = 0.4
  3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction
    Zn : HCl Or HCl : Zn
    1 : 2   1 : ½
  4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

        If all of the 0.5 moles of Zn were to be used in the reaction it would require
        2 x 0.5 = 1.0 moles of HCl for the reaction to go to completion.
        There are only 0.4 moles of HCl available which is less than the required 1.0 moles.

        If all of the 0.4 moles of HCl were to be used in the reaction it would require
        ½ x 0.4 = 0.2 moles Zn.
        There are 0.5 moles of Zn available which is more than the required 0.2 moles.

  5. The limiting reagent is the reactant that will be completely used up during the chemical reaction.
        There will be some moles of the reactant in excess left over after the reaction has gone to completion.

        The limiting reagent is HCl,
        all of the 0.4 moles of HCl will be used up when this reaction goes to completion.

        The reactant in excess is Zn,
        when the reaction has gone to completion there will be
        0.5 - 0.2 = 0.3 moles of Zn left over.

Masses of reactants given

Find the limiting reagent and the reactant in excess when 1.5 g of CaCO3 react completely with 0.73 g of HCl

  1. Write the balanced chemical equation for the chemical reaction

    CaCO3 + 2HCl → CaCl2 + CO2 + H2O

  2. Calculate the available moles of each reactant in the chemical reaction
    moles of CaCO3 = mass ÷ molar mass
    mass = 1.5 g
    molar mass = 40.08 + 12.01 + (3 x 16.00)
    = 100.09 g/mole

    moles of CaCO3 = 1.5 ÷ 100.09
    = 0.015 mol

    moles of HCl = mass ÷ molar mass
    mass = 0.73g
    molar mass = 1.008 + 35.45
    = 36.458g/mol

    moles HCl = 0.73 ÷ 36.458
    = 0.02 mol

  3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction
    CaCO3 : HCl Or HCl : CaCO3
    1 : 2   1 : ½

  4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

        If all of the 0.015 moles of CaCO3 were to be used in the reaction it would require
        2 x 0.015 = 0.03 moles of HCl for the reaction to go to completion.
        There are only 0.02 moles of HCl available which is less than the required 0.03 moles.

        If all of the 0.02 moles of HCl were to be used in the reaction it would require
        ½ x 0.02 = 0.01 moles of CaCO3.
        There are 0.015 moles of CaCO3 available which is more than the required 0.01 moles.

  5. The limiting reagent is the reactant that will be completely used up during the chemical reaction.
        There will be some moles of the reactant in excess left over after the reaction has gone to completion.

        The limiting reagent is HCl,
        all of the 0.02 moles of HCl will be used up when this reaction goes to completion.

        The reactant in excess is CaCO3,
        when the reaction has gone to completion there will be
        0.015 - 0.01 = 0.005 moles of CaCO3 left over.

Concentration and volume of solutions given

Find the limiting reagent and the reactant in excess when 100 mL of 0.2 mol L-1 NaOH aqueous solution react completely with 50 mL of 0.5 mol L-1 H2SO4 aqueous solution.

  1. Write the balanced chemical equation for the chemical reaction

    2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

  2. Calculate the available moles of each reactant in the chemical reaction:
    n = c x V
        n = moles of solute
        c = concentration of solution in mol L-1
        V = volume of solution in L
    n(NaOH) = c(NaOH) x V
    c(NaOH) = 0.2 mol L-1
    V = 100 mL = 100 x 10-3 L

    n(NaOH) = 0.2 x 100 x 10-3 = 0.02 mol

    n(H2SO4) = c(H2SO4) x V
    c(H2SO4) = 0.5 mol L-1
    V = 50 mL = 50 x 10-3 L

    n(H2SO4) = 0.5 x 50 x 10-3 = 0.025 mol

  3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction
    NaOH : H2SO4 Or H2SO4 : NaOH
    1 : ½   1 : 2

  4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

        If all of the 0.02 moles of NaOH were to be used in the reaction it would require
        ½ x 0.02 = 0.01 moles of H2SO4 for the reaction to go to completion.
        There are 0.025 moles of H2SO4 available which is more than the required 0.01 moles.

        If all of the 0.025 moles of H2SO4 were to be used in the reaction it would require
        2 x 0.025 = 0.05 moles of NaOH.
        There are only 0.02 moles of NaOH available which is less than the required 0.05 moles.

  5. The limiting reagent is the reactant that will be completely used up during the chemical reaction.
        There will be some moles of the reactant in excess left over after the reaction has gone to completion.

        The limiting reagent is NaOH,
        all of the 0.02 moles of NaOH will be used up when this reaction goes to completion.

        The reactant in excess is H2SO4,
        when the reaction has gone to completion there will be
        0.025 - 0.01 = 0.015 moles of H2SO4 left over.

Gas Volumes given

Find the limiting reagent and the reactant in excess when 45.42 L of CO(g) react completely with 11.36 L of O2(g) at STP (0oC or 273.15 K and 100 kPa)

  1. Write the balanced chemical equation for the chemical reaction

    2CO(g) + O2(g) → 2CO2(g)

  2. Calculate the available moles of each reactant in the chemical reaction
    moles of CO = V ÷ Vm
    At STP. 1 mole of gas
    has a volume of 22.71L

    moles of CO = 45.42 ÷ 22.71
        = 2 mol

    moles of O2 = V ÷ Vm
    At STP 1 mole of gas
    has a volume of 22.71L

    moles O2 = 11.36 ÷ 22.71
        = 0.5 mol

  3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction
    CO : O2 Or O2 : CO
    1 : ½   1 : 2

  4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

        If all of the 2 moles of CO were to be used in the reaction it would require
        ½ x 2 = 1 mole of O2 for the reaction to go to completion.
        There are 0.5 moles of O2 available which is less than the required 1 mole.

        If all of the 0.5 moles of O2 were to be used in the reaction it would require
        2 x 0.5 = 1 mole of CO.
        There are 2 moles of CO available which is more than the required 1 mole.

  5. The limiting reagent is the reactant that will be completely used up during the chemical reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion.

        The limiting reagent is O2,
        all of the 0.5 moles of O2 will be used up when this reaction goes to completion.

        The reactant in excess is CO,
        when the reaction has gone to completion there will be
        2 - 1 = 1 mole of CO left over.


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