Find the volume of carbon dioxide formed during the oxidation of 60 g of glucose in the human body.

Let’s write the solution based on the reaction equation:
С6Н12О5 + 6О2 = 6СО2 + 6Н2О + 38 ATP – the reaction of glucose oxidation in the body, aerobic decomposition, carbon dioxide is released;
Determine the molar mass of C5H12O6:
M (C6H12O6) = 12 * 6 + 12 + 16 * 6 = 180 g / mol;
Let’s calculate the number of moles of glucose according to the formula: Y (C5H12O6) = m (C6H12O5) / M (C5H12O6); Y (C5H12O6) = 60/180 = 0.3 mol;
Based on the data on the reaction equation, we will compose the proportion:
0.3 mol (C5H12O6) – X mol (CO2)
-1 mol – 6 hence, X mol (CO2) = 0.3 * 6/1 = 1.8 mol;
Let’s determine the volume of CO2 according to Avogadro’s law:
V (CO2) = 1.8 * 22.4 = 40.3 liters.
Answer: when glucose is oxidized, CO2 is released in a volume of 40.3 liters.