24.4 g of benzoic acid is converted to aniline by the Schmidt reaction (C6H5COOH + HN3 -> C6H5NH2 + CO2 + N2) with an overall yield of 80 %. Calculate the mass of aniline obtained in grams (C = 12, H = 1, N = 14, O = 16). Give the answer to one decimal place.

MolSynapse / Question bank / Aldehydes, Ketones and Carboxylic Acids / Carboxylic acids: acidity, preparation, derivatives, HVZ reaction

JEE Advanced levelNumerical answerDifficulty 3 of 3co214

24.4 g of benzoic acid is converted to aniline by the Schmidt reaction ( $C X_{6} H X_{5} COOH + HN X_{3} C X_{6} H X_{5} NH X_{2} + CO X_{2} + N X_{2}$ ) with an overall yield of 80 %. Calculate the mass of aniline obtained in grams (C = 12, H = 1, N = 14, O = 16). Give the answer to one decimal place.

Numerical answer type, answer in g, tolerance 0.05.

Answer: 14.9 g

Moles of benzoic acid = 24.4 / 122 = 0.200. Moles of aniline at 100 % yield = 0.200, and at 80 % yield = 0.160. Molar mass of aniline = 93 g/mol, so mass = 0.160 x 93 = 14.88 g, i.e. 14.9 g.

Practice Aldehydes, Ketones and Carboxylic Acids More on Carboxylic acids: acidity, preparation, derivatives, HVZ reaction

Previous questionNext question

24.4 g of benzoic acid is converted to aniline by the Schmidt reaction (CX6​HX5​COOH+HNX3​​CX6​HX5​NHX2​+COX2​+NX2​) with an overall yield of 80 %. Calculate the mass of aniline obtained in grams (C = 12, H = 1, N = 14, O = 16). Give the answer to one decimal place.

Answer: 14.9 g

More questions on Carboxylic acids: acidity, preparation, derivatives, HVZ reaction