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Math Help - Stoichiometry(Limiter, Percent Yield) Help

  1. #1
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    Stoichiometry(Limiter, Percent Yield) Help

    Hi everyone, I'm having trouble with the following questions
    Reaction: SrCl2+CuSO4 SrSO4 + CuCl2

    25.0 mL of .1 molar strontium chloride and 0.99 g of copper.

    Here are the Questions:
    1) Determine which reactant is limiting and what mass of the other reactant should remain after the reaction is complete.
    2) Determine the theoretical yield that was expected.
    3) Write a net ionic equation for this reaction.

    My Attempt at the question:

    Moles of SrCl2 = 0.0025 (0.025X0.1)

    Moles of Cu(SO4)
    = m/mm
    = 0.99/159.56
    =0.0062 mols

    This would make SrCl2 the limiter in the reaction. Is that right?

    SrSO4 is the solid that forms...
    1 mol CuSO4 reacts to make 1 mol SrSO4
    So the mols of excess reagent left over would be 0.0062-0.0025 = 0.0037

    m = n x mm
    = (0.0037 x 159.6)
    = 0.59g of the excess left over.

    Does that sound right?

    2) 1 mol of SrCl2 reacts to make 1 mol SrSO4
    0.0025 mol x
    x= 0.0025 mol
    m = n x mm
    = (0.0025x183.7)
    = 0.46 g of SrSO4

    3) No idea how to make the net ionic equation

    Sorry if my answers are a little unclear I'm not exactly sure if I'm doing the right thing. Thanks everyone

    Also, If you need me to clarify anything just lemme know
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  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
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    They all seem good to me. (I haven't checked the relative formula masses though)

    The ionic equation will be:

    SrCl_2 (aq) + CuSO_4 (aq) \rightarrow SrSO_4 (s) + CuCl_2 (aq)

    Write the state symbols. These will help you in such situations. Aqueous compounds will be in the form of ions, and solid compounds will remain as such.

    Sr^{2+} (aq)+ 2Cl^- (aq)+ Cu^{2+} (aq)+ SO_4\ ^{2-}(aq) \rightarrow SrSO_4 (s)+ Cu^{2+}(aq) + 2Cl^-(aq)

    Cross out the ions appearing on both sides:

    Sr^{2+} (aq)+SO_4\ ^{2-}(aq) \rightarrow SrSO_4(s)

    There you are!
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