Posted by **Anonymous** on Saturday, February 2, 2013 at 7:39pm.

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 -
**Devron**, Saturday, February 2, 2013 at 10:50pm
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 -
**Devron**, Saturday, February 2, 2013 at 11:23pm
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 -
**DrBob222**, Sunday, February 3, 2013 at 12:10am
Cr2O7^2- ==> 2Cr^3+ + 6e

a.

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 -
**Devron**, Sunday, February 3, 2013 at 12:56am
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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