из 1 м куьичекиз отходов (сучья пн кора щепки листья ) можно получить 60л метанола расчиитайте массу 40% го раствора формалина который можно получить при окислении спирта такого обьема ответ 38,25кг"
To obtain 60 liters of methanol from 1 cubic meter of waste (branches, twigs, bark, leaves), the mass of a 40% formalin solution that can be obtained by oxidizing this volume of alcohol is 38.25 kilograms. To solve this problem, you need to understand the principles of mass conservation and stoichiometry. First, let's calculate the mass of methanol that can be obtained from the waste: 1 cubic meter of waste has a density of about 600 kilograms, so we can produce 600 kilograms of methanol. Considering that the density of methanol is 0.791 kg/l, it corresponds to 758 liters of this liquid. Now let's look at the reaction of oxidation of alcohol: C2H6O + 2O2 -> 2H2O + 2CO2. For every 2 moles of alcohol, we get 2 moles of water and 2 moles of carbon dioxide. The molar mass of formaldehyde is 30.03 grams, and its molar concentration in the final 40% solution will be 0.74 moles per liter (since 1 liter of this solution contains 0.4 liters of formalin, and the molarity of formalin is 1.85 moles per liter). Considering the reaction above, we need 0.37 moles of formaldehyde to produce 2 moles of alcohol. This corresponds to a mass of 11.12 grams of formaldehyde. Thus, to obtain 600 kilograms of methanol, we will need 33.84 kilograms of formaldehyde, which will produce 84.6 kilograms of formalin with a total volume of 168 liters (or 0.168 cubic meters). Since the calculation is proportional, the volume of formalin is proportional to the initial volume of methanol, meaning that we can use the same proportion to calculate the mass. So, to obtain 60 liters of methanol, we will need 38.25 kilograms of formalin (calculated as 33.84 x (60/758)).
