In this experiment you will determine the percent by volume of alcohol in an alc

Question

In this experiment you will determine the percent by volume of alcohol in an alcoholic beverage. The alcohol is ethanol, C2H5OH. The alcohol and a lot of water will be removed from a sample of rum(40%alc.). The distillate, an aqueous ethanol solution will then be reacted with a measured, excess amount of standard potassium dichromate to oxidize the alcohol to acetic acid. 2C2H5OH + 2Cr2O72- + 16H+ ---> 3CH3COOH +4Cr3+ + 11H2O The excess dichromate is titrated with iron(II) ammonium sulfate. 6Fe2+ + Cr2O72- + 14H+ ---> 6Fe3+ + 2Cr3++7H2O

Use the following data to calculate the mass percent ethanol in a 50.532 g sample.
0.1956 M K2Cr2O7 added to solution B.

31.05 mL of 0.05432 M Fe2+ needed to titrate an aliquot of solution B.

a) mmol Fe2+ added to titrate sample.

b) mmol Cr2O72- in excess in titrated sample.

c) mmol Cr2O72- in excess in solution B.

d) mmol Cr2O72- added to solution B.

e) mmol Cr2O72- reacted with C2H5OH in solution B.

f) mmol C2H5OH in solution B.

g) mmol C2H5OH in solution A.

h) mass percent C2H5OH in original unknown beer sample.

Explanation / Answer

From the data

a) mmol Fe2+ added = 0.05432 M x 31.05 ml = 1.687 mmol

b) mmol Cr2O7^2- excess = 0.281 mmol

c) mmol Cr2O7^2- excess in B = 0.28 mmol

d) mmol Cr2O7^2- added = 0.1956 M x 100 ml = 19.56 mmol

[Note : check volume of solution B, here I have taken it as 100 ml]

e) mmol Cr2O7^2- reacted = 19.56 - 0.28 = 19.28 mmol

f) mmol C2H5OH in solution B = 19.28 mmol

g) mmol C2H5OH in solution A [solution A volume is not mentioned]

h) mass percent C2H5OH in original solution [depends on solution A volume]

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