If 6.46 L of gaseous ethanol reacts with 16.1 L O₂,what is the maximum volume of gaseous carbon dioxide produced? Assume that the temperature of the reactants and products is 425°C and the pressure remains constant at 1.00 atm. CH₃CH₂OH(g)+ 3O₂(g)→ 2CO₂(g)+ 3H₂O(g)

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The balanced chemical equation for the reaction of ethanol with oxygen is: $$C_2H_5OH(g) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(g)$$. According to the stoichiometry of the reaction, 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide. Given that the volumes of gases at constant temperature and pressure are directly proportional to the moles of gas (Avogadro's law), the volume of carbon dioxide produced can be calculated based on the limiting reactant. Since 6.46 L of ethanol would require $6.46 L \times 3 = 19.38 L$ of oxygen to fully react, but only 16.1 L of oxygen is available, oxygen is the limiting reactant. Therefore, the volume of carbon dioxide produced is $(16.1 L / 3) \times 2 = 10.7 L$, as it is produced in a 2:3 ratio with the limiting reactant oxygen.

If 646 L of Gaseous Ethanol Reacts with 16