(20 points The ethanol (FW 46.070) in a 20.00 mL aliquot was distilled into 100.
ID: 508879 • Letter: #
Question
(20 points The ethanol (FW 46.070) in a 20.00 mL aliquot was distilled into 100.0 mL of 0.05151 M K2Cr207 (FW 294.19). Heating completed the oxidation of the alcohol to acetic acid (FW 60.054), following which the excess dichromate was titrated with 34.42 mL of 0.02497 MFe (FW 55.847). a) Write the balanced chemical equation for the reaction which occurs between the ethanol and the dichromate. b) Write the balanced chemical equation for the titration reaction c) Determine the mass of ethanol in the brandy sample.Explanation / Answer
During distillation Ethanol is oxidized by dichromate.
Balanced chemical reaction:
Write the reduction half reactions.
CH3COOH +4H+ + 4e- ----> CH3CH2COOH + H2O
Cr2O72- + 14H+ + 6e- ----> 2Cr3+ + 7H2O
We have to balance the electron.
So multiply 3 with first equation and 2 with second equation to make 12 electron for both
3CH3COOH +12H+ + 12e- ----> 3CH3CH2COOH + 3H2O
2Cr2O72- + 28H+ + 12e- ----> 4Cr3+ + 14H2O
Subtract first equation from second equation.
3CH3CH2OH + 2Cr2O72-+16H+ ----> 3CH3COOH + 4Cr3+ + 11H2O
b)
Titration is between dichromate and Fe2+
Oxidation of Fe2+ by dichromate.
Write the reduction half reaction.
Fe2+ + e- ---> Fe3+
Cr2O72- + 14H+ + 6e- ----> 2Cr3+ + 7H2O
Multiply 6 with first equation.
6Fe2+ + 6e- ---> 6Fe3+
After subtraction we have
Cr2O72- + 14H+ + 6Fe2+ ----> 2Cr3+ + 7H2O + 6Fe3+
c)
Mass of ethanol
Number of moles of dichromate initially is
= 100 mL x 0.05151 mmol/mL
= 0.005151 mol
Number of moles of Fe2+ required to reduce dichromate
= 34.42 mL x 0.02497 mmol /mL
= 8.59 x 10-4 mol
From the equation
Cr2O72- + 14H+ + 6Fe2+ ----> 2Cr3+ + 7H2O + 6Fe3+
One mole of Cr2O72- requires six mol Fe2+
1 mol Cr2O72- = 6 mol Fe2+
1/6 mol Cr2O72- = 1 mol Fe2+
8.59 x 10-4 mol Fe2+ = 8.59 x 10-4 (1/6)ol Cr2O72-
= 1.432 x 10-4 mol of Cr2O72-
The above is the number of moles of Cr2O72- remaining after reacting with ethanol.
Number of moles of Cr2O72- reacted with ethanol is
= 0.005151 - 1.432 x 10-4
= 0.00500 mol
3CH3CH2OH + 2Cr2O72-+16H+ ----> 3CH3COOH + 4Cr3+ + 11H2O
From the above equation
3 mol of ethanol requires 2 moles of Cr2O72-
3 mol ethanol = 2 mol of Cr2O72-
1 mol of of Cr2O72- = 3/2 mol of ethanol
0.00500 mol of Cr2O72 = 0.00500 (3/2) mol of ethanol
=0.007511 mol
20mL contains 0.007511 moles.
100mL contains 0.007511 x 5 = 0.03755 moles.
Mass of ethanol is
= 0.03755 x 40.070
Mass of ethanol is 1.73 g
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