Calculate the mass of glucose that has undergone alcoholic fermentation

Calculate the mass of glucose that has undergone alcoholic fermentation, if the same amount of carbon (IV) oxides has been released as it is formed by the complete combustion of 16 g of methyl alcohol.

Given:
m (CH3OH) = 16 g

To find:
m (C6H12O6) -?

Decision:
1) С6Н12О6 => 2С2Н5ОН + 2СО2 ↑;
2СН3ОН + 3 О2 => 2СО2 ↑ + 4 Н2О;
2) M (CH3OH) = Mr (CH3OH) = Ar (C) + Ar (H) * 4 + Ar (O) = 12 + 1 * 4 + 16 = 32 g / mol;
M (C6H12O6) = Mr (C6H12O6) = Ar (C) * 6 + Ar (H) * 12 + Ar (O) * 6 = 12 * 6 + 1 * 12 + 16 * 6 = 180 g / mol;
3) n (CH3OH) = m (CH3OH) / M (CH3OH) = 16/32 = 0.5 mol;
4) n (C6H12O6) = n (CO2) / 2 = n (CH3OH) / 2 = 0.5 / 2 = 0.25 mol;
5) m (C6H12O6) = n (C6H12O6) * M (C6H12O6) = 0.25 * 180 = 45 g.

Answer: The mass of С6Н12О6 is 45 g.