4. In this experiment ethanol, CH3CH2OH, is oxidized by chromic acid, H2CrO4 a)

Question

4. In this experiment ethanol, CH3CH2OH, is oxidized by chromic acid, H2CrO4 a) Potassium dichromate, KCr20, (aq) reacts with sulfuric acid, H SO4 (aq) to produce chromic acid H2CrO4 (aq) according to the following reaction. 2 H30' (aq) + Cr2072-(aq) ? 2 Hfro, (aq) + H2O (1) The chromic acid H2CrO4 (aq) is made by adding 1.00 mL of 0.0196 M potassium dichromate, KCr20, (aq), to 10.0 mL of 3.9 M sulfuric acid, H2SO4 (aq). Calculate the concentration of the chromic acid. 10.0 uL of CH, CH2OH is added to 2.00 mL of H2CrO, (aq). Calculate the concentration of CH3CH2OH in the resulting solution given that the density of ethanol is 0.789 g/ml b) c) Ethanol is oxidized by the chromic acid according to the following reaction. H,O (aq) +3 CH,CH2OH (aq) +2 H2CrO, (aq) +3 CH,CHO (aq) +2Cr3+ (aq)+ 14 H20 0) Calculate the concentration of ethanol after reaction goes to completion and the percent change in the concentration of ethanol.

Explanation / Answer

Ans. #4a. Moles of K-dichromate = Molarity x Vol. of soln. in liters

                                                = 0.0196 M x 0.001 L = 0.0000196 mol

#Moles of H2SO4 = 3.9 M x 0.010 L = 0.039 mol

# Since moles of K2Cr2O7 is far less than its stoichiometry ratio to H2SO4, it is the limiting reactants.

# The reaction follows the stoichiometry of limiting reactant.

# Following the stoichiometry of balanced reaction, 1 mol K2Cr2O7 forms 1 mol chromic acid.

So,

            Moles of H2CrO4 formed = 0.0000196 mol

# Total vol of reaction mixture = 1.0 mL + 10.0 mL = 11.0 mL = 0.011 L

Now,

            [H2CrO4] = 0.0000196 mol / 0.011 L = 0.00178 M

#4b. Mass of ethanol added = Vol x Density = 0.010 mL x (0.789 g/ mL) = 0.00789 g

Moles of ethanol = 0.00789 g / (46.06904 g/mol) = 1.7126 x 10-4 mol

# Total vol of reaction mixture = 0.010 mL + 2.00 mL = 2.010 mL = 0.002010 L

Now,

            Initial [C2H5OH] = 1.7126 x 10-4 mol / 0.002010 L = 0.0852 M

#4c. We have,

            Moles of ethanol = 0.00017126 mol

            Moles of chromic acid = [H2CrO4] in #4a x Vol. of soln. taken in liters

                                                = 0.00178 M x 0.002 L = 0.00000356 mol

# Following stoichiometry of balanced reaction, 3 mol ethanol reacts with 2 mol chromic acid. Since moles of H2CrO4 is far less than its stoichiometry ratio to ethanol, it is the limiting reactants. The reaction follows the stoichiometry of limiting reactant.

So,

            Moles of ethanol consumed = (3/2) x Moles of H2CrO4

                                                            = (3/2) x 0.00000356 mol

                                                            = 0.00000534 mol

            Remaining moles of ethanol = Initial moles – Moles consumed

                                                            = 0.00017126 mol - 0.00000534 mol

                                                            = 0.00016592 mol

# Final [C2H5OH] = 0.00016592 mol / 0.002010 L = 0.0825 M

# Change in [C2H5OH] = Final – Initial concertation = 0.0825 M – 0.0852 M

                                                = -0.0027 M

# Change = (Change in concentration / Initial concertation) x 100

                        (-0.0027 M / 0.0852 M) x 100

                        = -3.17 %

The –ve sing indicates decrease in ethanol concertation.

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