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

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

Idon't know hope someone you