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)

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 = ?

To determine the percentage of aluminum (% Al) in the sample, we need to calculate the amount of iron reacting with the potassium dichromate and then compare it to the total mass of the sample.

1. Calculate the amount of Fe(II) reacting with potassium dichromate:
- From the reaction equation, we know that 1 mole of Fe(II) reacts with 1 mole of Cr2O7(-2) (dichromate ion).
- The molar ratio between Fe(II) and dichromate is 1:1.
- We have the volume and concentration of the potassium dichromate solution used, so we can calculate the number of moles of dichromate used in the titration:

Moles of dichromate = Volume (L) × Concentration (mol/L)
= 0.01436 L × 0.0845 mol/L

2. Determine the moles of Fe(II) reacting with the dichromate:
- Since the molar ratio between Fe(II) and dichromate is 1:1, the moles of Fe(II) reacting would be the same as the moles of dichromate used.

3. Calculate the molar mass of Fe(II) (iron):
- The molar mass of iron (Fe) is 55.845 g/mol.

4. Determine the mass of Fe(II) in the sample:
- Multiply the moles of Fe(II) by the molar mass of Fe(II):

Mass of Fe(II) = Moles of Fe(II) × Molar mass of Fe(II)
= Moles of Fe(II) × 55.845 g/mol

5. Determine the total mass of the sample:
- The total mass of the sample is given as 2.764 g.

6. Calculate the mass of aluminum (Al) in the sample:
- Subtract the mass of Fe(II) from the total mass of the sample to find the mass of Al:

Mass of Al = Total mass of the sample - Mass of Fe(II)

7. Calculate the percentage of Al in the sample:
- Divide the mass of Al by the total mass of the sample and multiply by 100:

% Al = (Mass of Al / Total mass of the sample) × 100

By following these steps, you can determine the percentage of Al in the sample.

To find the percentage of aluminum (% Al) in the sample, we need to first determine the number of moles of iron (Fe) reacted with the potassium dichromate.

1. Calculate the number of moles of potassium dichromate used:
Number of moles = Molarity * Volume
Number of moles = 0.0845 M * 0.01436 L = 0.00121122 mol

2. Determine the number of moles of Fe reacted:
From the balanced equation, we know that the molar ratio between Fe and potassium dichromate is 1:6.
Therefore, the number of moles of Fe is:
Number of moles of Fe = 0.00121122 mol * (1 mol Fe / 6 mol K2Cr2O7) = 0.00020187 mol

3. Find the molar mass of Fe:
The molar mass of Fe is approximately 55.845 g/mol.

4. Calculate the mass of Fe in the sample:
Mass of Fe = Number of moles of Fe * Molar mass of Fe
Mass of Fe = 0.00020187 mol * 55.845 g/mol = 0.0112678 g

5. Subtract the mass of Fe from the total sample mass to determine the mass of Al:
Mass of Al = Total sample mass - Mass of Fe
Mass of Al = 2.764 g - 0.0112678 g = 2.7527322 g

6. Calculate the % Al in the sample:
% Al = (Mass of Al / Total sample mass) * 100
% Al = (2.7527322 g / 2.764 g) * 100 = 99.6%

Therefore, the sample is approximately 99.6% aluminum (% Al).