analytical chemisty
posted by Anonymous .
Consider the standardization of a solution of K2Cr2O7 with iron metal according to the following:
Reaction: Fe + Cr2O72 → 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?

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 
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. 
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. 
Okay, my setup was correct then. I just didn't show how I got 43.09g.
B is what I was worried about!!!!!!