Under the action of nitrous acid on 1.0 g of a mixture containing alanine, 92.4 cm3

univerkov | August 7, 2021 | education

Under the action of nitrous acid on 1.0 g of a mixture containing alanine, 92.4 cm3 of nitrogen was formed. Calculate the percentage of alanine in the test mixture.

Let’s find the amount of nitrogen substance.

n = V: Vn.

92.4 cm3 = 0.092 l.

n = 0.092 L: 22.4 L / mol = 0.00411 mol.

CH3CH (NH2) COOH + HNO2 → CH3CH (OH) COOH + N2 + H2O.

n (N2) = n (CH3CH (NH2) COOH) = 0.0041 mol.

Find the mass CH3CH (NH2) COOH.

M (CH3CH (NH2) COOH) = 89 g / mol.

m = n × M.

m = 89 g / mol × 0.0041 mol = 0.36 g.

Find the mass fraction of alanine in the mixture.

w = (0.36 g: 1 g) × 100% = 36%.

Answer: 36%.

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