Calculate the volume of carbon monoxide (IV) formed during the oxidation of 0.25 mol of sucrose.

C12H22O11 + 12O2 = 12CO2 + 11H2O.

Taking into account the stoichiometric coefficients, we determine that 12 mol of carbon dioxide is formed from 1 mole of sucrose. Let’s make the proportion and find the amount of carbon monoxide substance formed from 0.25 mol of sucrose:

0.25 mol – x mol

1 mol – 12 mol.

x = 0.25 * 12 = 3 mol. We will find the volume of carbon monoxide based on the fact that 1 mol of gas at normal conditions. takes up a volume of 22.4 l:

V = 22.4 * 3 = 67.2 l.