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analytical chemisty

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Consider the standardization of a solution of K2Cr2O7 with iron metal according to the following:

Reaction: Fe + Cr2O7-2 → Fe +3 + Cr+3a.

How many grams of K2Cr2O7 are required to prepare 500 ml of a 0.13 N solution?

What is the exact normality of the solution prepared in part a if 0.1376 g of iron metal were exactly reacted with 48.56 ml of the solution?

  • Analytical Chemisty -

    0.13N= # of equivalents/500 x10^-3 L, solving for # of equivalents

    # of equivalents in 0.13N=(0.13N)(500 x 10^-3L)

    # of equivalents in 0.13N= g of solute/49.03g, solve for grams of solute,

    (49.03g, solve )*(# of equivalents in 0.13N)= g of solute

  • Analytical Chemisty -

    I'm not sure if you are asking for the normality of Fe, but if you are use

    NV=NV, and solve for N

    N=[(500x10^-3 l)(0.13N)]/48.86 x 10^-3 L

    N= 1.33N

    Hopefully Dr.Bob222 checks this and corrects it if it is wrong.

  • analytical chemisty -

    Cr2O7^2- ==> 2Cr^3+ + 6e

    mL x N x m.e.w. = grams.
    500 mL x 0.13 x (294.18/6000) = grams = about 3.2 grams K2Cr2O7.

    I won't do b for you; Fe doesn't react with Cr2O6^2-. With Fe^2+ yes but not Fe metal.

  • analytical chemisty -

    Okay, my setup was correct then. I just didn't show how I got 43.09g.

    B is what I was worried about!!!!!!

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