How much CO2 will be released during alcoholic fermentation of glucose m = 270 kg?

Let’s make the equation:
С6Н12О6 = 2С2Н5 – ОН + 2СО2 – the reaction of glucose fermentation occurs with the participation of yeast, ethyl alcohol is released.
Let’s calculate the molar mass of C6H12O6: M (C6H12O6) = 12 * 6 + 12 + 16 * 6 = 180 g / mol;
Let us determine the number of moles of С6Н12О6, if the mass is known:
Y (C6H12O6) = m; (C6H12O6) / M (C6H12O6); Y (C6H12O6) = 27000/180 = 150 mol
Let us determine X mol of CO2 by the proportion:
150 mol (С6Н12О6) – х mol (СО2)
X mol (CO2) = 150 * 2/1 = 300 mol;
We calculate the volume of CO2 according to Avogadro’s law:
V (CO2) = Y (CO2) * 22.4; V (CO2) = 300 * 22.4 = 6720 l.
Answer: during alcoholic fermentation of glucose, CO2 is released with a volume of 6720 liters.