The compete oxidation of 2g of an aldehyde A by acidified potassium chromate produces 2.1g of carboxylic acid B which is dissolved in 50 ml of distilled water .

The obtained solution reacts with exactly 1.05 g of solid lime to form salt C
a) Write the equation of the reaction B with the lime.
b) Calculate the molar mass of acid B. Deduce its molecular formula

1 answer

A + [O] ==> B + Ca(OH)2 ==> C
2g.........2.1....1.05.........

Ca(OH)2 + 2XCOOH ==> Ca(XCOO)2 + 2H2O
mols Ca(OH)2 = grams/molar mass = approx 0.014 but you need to do that more accurately.
mols XCOOH must be twice that or about 0.028.
mols = grams/molar mass or
molar mass = grams/mols = approx 2.1/about 0.028 = 74.
If the molar mass is 74, the COOH part of that must be 45 which leaves for the X part of it 74-45 = 29. That's probably CH3CH2 so I would think B is CH3CH2COOH or propanoic acid.