How much carbon dioxide will be released when 64 g of methyl alcohol is burned?

2CH3OH + 3O2 -> 2CO2 + 4H2O – methanol burns to form carbon dioxide and water.
n = 2 moles – for the reaction equation we have 2 moles of methanol and 2 moles of carbon dioxide.
M = 32 g / mol is the molar mass of methanol, for ny it will have a volume of 22.4 liters, but we have two moles, therefore, twice as much: m = 64 g, V = 44.8 liters. Now add up the proportion: 64 g of CH3OH – 44.8 l of CO2, and 64 g of CH3OH – X l of CO2; V (CO2) = 64 * 44.8 / 64 = 44.8 liters.
Answer: the volume of carbon dioxide that was released during the combustion of 64 g of methyl alcohol = 44.8 liters.