# Chemistry

A 2.764 g sample containing only iron and aluminum metal was treated with acid to form Al(+3) and Fe(+2). The sample was then titrated with a potassium dichromate solution (only the Fe(II) will react with the dichromate). The sample was found to be equivalent to 14.36 mL of 0.0845 M potassium dichromate solution. What is the % Al in the sample?

Fe(+2) + Cr2O7(-2) ---------> Cr(+3) + Fe(+3)

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1. Balance the equation between Fe(II) and Cr2O7^2-.
Frankly, I am not familiar with the term "equivalent to xx mL of yy M" something else. I assume that means that xx mL of yyM was used to titrate it to the equivalence point. If that is it, then
Convert mols Cr2O7^2- to mols fe(II) using the coefficients in the balanced equation.
Then g Fe = mols Fe x atomic mass Fe.

mass sample-grams Fe = grams Al.
Then (g Al/mass sample)*100 = %Al

Or you could do
%Fe = (g Fe/mass sample)*100 = ?
and % Al = 100-%Fe.
mols Car2O7^2- = M x L = ?

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