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March 27, 2017

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what volume (in L) of 0.150 M KCl solution is required to completely react with 0.150 L of a 0.175 M Pb(NO3)2 solution according to the following balanced chemical equation?

2KCl (aq) + Pb(NO3)2 (aq) --> PbCl2 (s) + 2 KNO3 (aq)

  • chemistry - ,

    1. Calculate moles Pb(NO3)2. mols = M x L.

    2. Using the coefficients in the balanced equation, convert moles Pb(NO3)2 to moles KCl.

    3. Convert mols KCl to L. M = mols/L.

  • chemistry - ,

    okay so the solution would be.

    0.150 M KCl x 0.175 M Pb(NO3)2/ 1 M KCL x 0.150 L/0.175 M Pb(NO3)2 = 0.023 L?

  • chemistry - ,

    No. Don't try to do it all at once. Do it in steps as I've suggested.
    0.15 x 0.175 will give you the correct mols Pb(NO3)2 but your conversion factor does not convert to moles KCl. There are 2 moles KCl/1 mole Pb(NO3)2 according to your balanced equation. I don't know what the last term means.

  • chemistry - ,

    my teacher insists that we do it that way, i don't really understand it but i try!

    so it would come out to be 0.026 moles Pb(NO3)2. Then, converting Pb(NO3)2 into KCl it would come out to be 0.052 Moles KCl. Then divide 0.052 by 0.150 which would equal 0.15 L?

  • chemistry - ,

    you are right but when you divide 0.052 by 0.150 the answer will be 0.35 L.

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