The oxidation reaction of the amino acid alanine (NH2CHCH3COOH) is described as:

Question

The oxidation reaction of the amino acid alanine (NH2CHCH3COOH) is described as: 2 NH 2CHCH 3COOH (s ) + 9 O 2 (g ) -> N 2 (g ) + 6 H 2O (l ) + 6 CO 2 (g ) Calculate the volume of all gases evolved by the complete oxidation of 0.25 g of the amino acid alanine (Malanine = 88.07 g/mol) if the products are liquid water, nitrogen gas, and carbon dioxide gas and the total pressure is 1.00 atm and T = 310.0 K. The oxidation reaction of the amino acid alanine (NH2CHCH3COOH) is described as: 2 NH 2CHCH 3COOH (s ) + 9 O 2 (g ) -> N 2 (g ) + 6 H 2O (l ) + 6 CO 2 (g ) Calculate the volume of all gases evolved by the complete oxidation of 0.25 g of the amino acid alanine (Malanine = 88.07 g/mol) if the products are liquid water, nitrogen gas, and carbon dioxide gas and the total pressure is 1.00 atm and T = 310.0 K. 2 NH 2CHCH 3COOH (s ) + 9 O 2 (g ) -> N 2 (g ) + 6 H 2O (l ) + 6 CO 2 (g ) Calculate the volume of all gases evolved by the complete oxidation of 0.25 g of the amino acid alanine (Malanine = 88.07 g/mol) if the products are liquid water, nitrogen gas, and carbon dioxide gas and the total pressure is 1.00 atm and T = 310.0 K.

Explanation / Answer

2 NH 2CHCH 3COOH (s ) + 9 O 2 (g ) -> N 2 (g ) + 6 H 2O (l ) + 6 CO 2 (g )

0.25 g of the amino acid alanine

Molar mass of alanine = 88.07 g/mol

No of moles = 0.25/88.07 = 0.002838651 moles

Now, no of moles of gases( N2 + CO2), n = ((1+6)/2) *  0.002838651 moles

= 0.0099352785 moles

Now,

PV = nRT

where, R = 0.08206 L atm mol

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