A 2.764 g sample containing only iron and aluminum metal was treated with acid t

Question

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 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

Explanation / Answer

mass of Fe = mol *MW

the actual equation

6Fe+2 + CrO7-2 + 15H+ = 6Fe+3 + 2Cr+3+7H2O

then 6 mol of Fe+ per mol of chromate

then

mol of chromte = MV = 14.36*0.0845 = 1.21342 molof CrO7-2

then

6x is Fe+2

1.21342*6 = 7.28052 mmol of Fe+

then mass = mol*MW = (7.28052*10^-3)(55.5) = 0.4040 g of Fe

then

mass of Al =  2.76-0.4040 = 2.356 gof Al

% Al = 2.356/2.764 *100 = 85.2 % of Al

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