Using the level of the chemical reaction C6H12O6-2C2H5OH + 2CO2, calculate the volume of carbon monoxide released during alcoholic fermentation of glucose if 230 g of ethyl alcohol is obtained.
1. Determine the amount of ethanol, if its mass: m (C2H5OH) = 230g; M (C2H5OH) = 46 g / mol.
Y (C2H5OH) = m (C2H5OH) / M (C2H5OH): Y (C2H5OH) = 230/46 = 5 mol.
2. Taking into account the data on the reaction equation, we will compose the proportion:
X mol (C6H12O6) – 5 mol (C2H5OH)
X mol (C6H12O6) = 1 * 5/2 = 2.5 mol.
3. Determine the number of moles of CO2 based on the proportion:
2.5 mol (C6H12O6) – X mol (CO2)
X mol (CO2) = 2.5 * 2/1 = 5 mol.
4. Let’s calculate the volume of CO2 according to Avogadro’s law:
V (CO2) = 5 * 22.4 = 112 liters.
Answer: during alcoholic fermentation of glucose, 112 liters were released. carbon dioxide.
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