Make the equation for the reaction proceeding according to the scheme: starch-glucose-ethanol

Make the equation for the reaction proceeding according to the scheme: starch-glucose-ethanol-carbon monoxide (IV) -starch. What amount of carbon monoxide (IV) was released in the third stage if starch with a mass of 243 g was taken?

Given:
m (starch) = 243 g

To find:
n (CO2) -?

Decision:
1) (C6H10O5) n + nH2O = (H2SO4) => nC6H10O6;
C6H10O6 = (alcohol fermentation) => 2C2H5OH + 2CO2;
2C2H5OH + 7O2 => 4CO2 + 6H2O;
6nCO2 + 5nH2O = (photosynthesis) => (C6H10O5) n + 6nO2;
2) M (starch) = (162n) g / mol;
3) n (starch) = m (starch) / M (starch) = 243 / 162n = (1.5 / n) mol;
4) n (C6H10O6) = n (starch) * n = (1.5 / n) * n = 1.5 mol;
5) n (C2H5OH) = n (C6H10O6) * 2 = 1.5 * 2 = 3 mol;
6) n (CO2) = n (C2H5OH) * 2 = 3 * 2 = 6 mol.

Answer: The amount of CO2 is 6 mol.