Question Solved1 Answer Help QUESTION 5 5.1 Balance the following chemical reaction in basic solution: Mno* + N2H4 MnO2 + N2 [5] 5.2 A 5.00-ml sample of brandy was diluted to 1.000 L in a volumetric flask. The ethanol (C2H5OH) in a 25.00-ml aliquot of the diluted soln was distilled into 50.00 ml of 0.020 M K Cr20., and oxidized to acidic acid with heating. After cooling, 20.00 mL of 0.1253 M Fe2+ were pipetted into the flask. The excess Fet was then titrated with 7.46 mL of the standard K2Cr20, to a diphenylamine sulfonic acid end point. [3] (a) Write a balanced redox reaction of ethanol with the dichromate ion Calculate the percent (w/v) C2H5OH (46.07 g/mol) in the brandy [6] (b)

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Transcribed Image Text: QUESTION 5 5.1 Balance the following chemical reaction in basic solution: Mno* + N2H4 MnO2 + N2 [5] 5.2 A 5.00-ml sample of brandy was diluted to 1.000 L in a volumetric flask. The ethanol (C2H5OH) in a 25.00-ml aliquot of the diluted soln was distilled into 50.00 ml of 0.020 M K Cr20., and oxidized to acidic acid with heating. After cooling, 20.00 mL of 0.1253 M Fe2+ were pipetted into the flask. The excess Fet was then titrated with 7.46 mL of the standard K2Cr20, to a diphenylamine sulfonic acid end point. [3] (a) Write a balanced redox reaction of ethanol with the dichromate ion Calculate the percent (w/v) C2H5OH (46.07 g/mol) in the brandy [6] (b)
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Transcribed Image Text: QUESTION 5 5.1 Balance the following chemical reaction in basic solution: Mno* + N2H4 MnO2 + N2 [5] 5.2 A 5.00-ml sample of brandy was diluted to 1.000 L in a volumetric flask. The ethanol (C2H5OH) in a 25.00-ml aliquot of the diluted soln was distilled into 50.00 ml of 0.020 M K Cr20., and oxidized to acidic acid with heating. After cooling, 20.00 mL of 0.1253 M Fe2+ were pipetted into the flask. The excess Fet was then titrated with 7.46 mL of the standard K2Cr20, to a diphenylamine sulfonic acid end point. [3] (a) Write a balanced redox reaction of ethanol with the dichromate ion Calculate the percent (w/v) C2H5OH (46.07 g/mol) in the brandy [6] (b)
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Answr 5.1: Step  1 : seperate half reaction  MnO4- -->  MnO2 N2H4 -->N2  Now we will balance oxygen by adding water as N an Mn are balanced  MnO4- -->  MnO2  + 2H2O N2H4 -->N2  Now we will balance hydrogen by adding proton MnO4- +4H+ -->  MnO2  + 2H2O N2H4 -->N2  +4H+ Balance the charge with electon MnO4- +4H+ + 3e---->  MnO2  + 2H2O N2H4 -->N2  +4H+ + 4e-  multiplying rection by  4 to first an 3 to secon reacton  4MnO4- +16H+ + 12e---->  4Mn ... See the full answer