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    • #27 - Stoichiometry Calculations of
    Double-Displacement Reactions

    If I wanted to make something out of two compounds by double-displacement how would I know how much I get?

    If I have 20 invitations and only 15 envelopes, I can only make 15 letters to mail out for a party.

    That means that the envelopes are the limiting reactant, the reactant in a balanced equation, the limiting reactant that controls how much (or many) product is made.

    Now, lets go to a real balanced equation:

    NaF + LiOH -> NaOH + LiF

    Since LiF is insoluble, this is the final product

    There are 3 grams of NaF, but only 2 grams of LiOH, lets convert those into moles:

    3 grams NaF | 1 mole NaF_____ = .07145 moles NaF
    | 41.99 grams NaF

    2 grams LiOH | 1 mole LiOH_____ = .08351 moles LiOH
    | 23.949 grams LiOH

    Since there�s less NaF than LiOH, and the mole ratio is 1to1(since the coefficients are both one), NaF is the limiting reactant. The amount of NaF(in moles) determines how much product is made, since NaF will run out before the LiOH just like the envelops will run out before the invitations.

    - Use the limiting reactant in figuring out the amount of a product

    -So lets find out the amount of product!

    3 grams NaF | 1 mol NaF _________ | 1 mol LiF__ | 25.9394 g LiF * = 1.85 g LiF
    | 41.9882 grams NaF | 1 mol NaF | 1 mol LiF

    The amount of moles does not determine the limiting reactant!

    To make things easier, there�s a 5 step process in which you can calculate the limiting reactant and the amount of the product.

    1. Make and balance an Equation
    2. Get moles of reactants
    3. Divide moles by coefficients
    4. The reactant with smallest amount from step 3 is the limiting reactant
    5. Use the limiting reactant to find grams of product

    Here�s the process in action:

    Kelly the chemist has 3 grams of Aluminum Nitrate(Al(NO3)3) and 4 grams of Sodium Chloride(NaCl), how much Aluminum Chloride will Kelly make?

    1. Make and balance an equation:

    Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3

    2. Get moles of reactants:

    3 grams Al(NO3)3 | 1 mole Al(NO3)3_____ = .01818 moles Al(NO3)3
    | 165.01 grams Al(NO3)3

    4 grams NaCl | 1 mole NaC1_____ = .06845 moles NaCl
    | 58.44 grams NaCl

    3. Divide moles by coefficients:

    .01818 moles Al(NO3)3 = .01818
    1

    .06845 moles NaCl = .02282
    3

    4. The reactant with smallest amount from step 3 is the limiting reactant:

    .01818 < .02282
    Al(NO3)3 is the limiting reactant

    5. Use the limiting reactant to find grams of product

    3 grams Al(NO3)3 | 1 mole Al(NO3)3____ | 1 mole AlCl3____ | 133.33 grams AlCl3 =2.4 grams AlCl3
    | 165.01 grams Al(NO3)3 | 1 mole Al(NO3)3 | 1 mole AlCl3

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