Question

Calculate the volume of all gases evolved by the complete oxidation of 0.380 g of the amino acid alanine

Answer 1

To calculate the volume of all gases evolved by the complete oxidation of 0.380 g of the amino acid alanine, determine the molar mass of alanine and use stoichiometry to find the volume of the gases produced.

Step-by-step explanation:

To calculate the volume of all gases evolved by the complete oxidation of 0.380 g of the amino acid alanine, we need to determine the molar mass of alanine and then use stoichiometry to find the volume of the gases produced.

First, calculate the molar mass of alanine (C3H7NO2) using the atomic masses of carbon, hydrogen, nitrogen, and oxygen. The molar mass of alanine is found to be 89.09 g/mol.

Next, use the balanced equation for the complete oxidation of alanine: C3H7NO2 + 4O2 → 3CO2 + 4H2O + 2N2. The stoichiometry of the equation tells us that for every 1 mole of alanine reacted, 3 moles of carbon dioxide (CO2) are produced.

Use the molar mass of CO2 (44.01 g/mol) and the given mass of alanine (0.380 g) to calculate the moles of alanine. Then, use the stoichiometry to calculate the moles of carbon dioxide produced. Finally, convert the moles of carbon dioxide to volume using the ideal gas law equation: PV = nRT. Substitute the values for pressure, temperature, and moles to find the volume of the gases evolved.

Answer 2

The products of oxidation of alanine are carbon dioxide, water and nitrogen. The substances that would be in the gas phase are carbon dioxide and nitrogen. The balanced chemical reaction is:

4 CH3CH(NH2)COOH + 15 O2 ---> 12 CO2(g) + 14 H2O(l) + 2 N2(g)

To determine the volume of these gases we calculate the number of moles produced. We do as follows:

moles alanine = 0.380 g ( 1 mol / 89.09 g alanine ) = 4.27x10^-3 mol

moles CO2 = 4.27x10^-3 mol CH3CH(NH2)COOH ( 12 mol CO2 / 4 mol CH3CH(NH2)COOH ) = 0.013 mol CO2

moles N2 = 4.27x10^-3 mol CH3CH(NH2)COOH ( 2 mol N2 / 4 mol CH3CH(NH2)COOH ) = 0.0021 mol N2

moles gas = 0.013 + 0.0021 = 0.015 moles

Assuming at STP, 1 mol of gas is equal 22.4 L

Volume = 0.015 ( 22.4 L / 1 mol ) = 0.34 L