Oxidation of aldehydes and ketones

Exercise 5

The catalytic oxidation is carried on $35\;cm^3 $ of ethanol ($\rho=790\frac{kg}{m^3} $). During this reaction, the alcohol is converted into an aldehyde and a portion thereof to the carboxylic acid. 1) On the half of the resulting mixture is reacted Fehling's solution. What is the precipitate obtained and how many moles of the precipitate obtained for one mole of reagent ? 2) It was found that the precipitate present after washing and drying had a mass of $28.6\;g$. Calculate the amount of aldehyde which was provided in the liquid mixture by the oxidation of alcohol. 3) On the other half of the mixture, an excess of magnesium is made to react. Calculate the volume of gas obtained at $26.85^oC$ and $1.00\;atm$

1) Precipitate: $Cu_2O$ One mole of copper(I) oxide is obtained from one mole of aldehyde. 2) naldehyde which reacted = $n_{Cu_2O}$ $=$ $\frac{28.6}{127+16}$ $=$ $0.200\; mol$ naldehyde in the final mixture = $0.400\; mol$ ninitial ethanol = $\frac{35\cdot 0.79}{46}$ $=$ $0.600\; mol$ = nethanal+ethanoic acid obtained nethanoic acid obtained = $0.600-0.400$ $=$ $0.200\; mol$ 3) nethanoic acid in the second half = $0.100 \; mol$ One mole of acid produces half a mole of hydrogen gas: nhydrogen = $0.050 \; mol$ Vhydrogen = $\frac{0.050 \cdot 0.082\cdot 300 }{1}$ $=$ $\frac{0.050 \cdot 0.082\cdot 300 }{1}$ $=$ $1.23\; L$