Final answer:
The volume of carbon dioxide produced at STP from the reaction of 0.900 moles of ethyl alcohol with 1.42 moles of oxygen is approximately 40.3452 liters, determined by the stoichiometric ratio of the reactants and the molar volume of a gas at STP.
Step-by-step explanation:
To determine the volume of carbon dioxide produced from the reaction of oxygen with ethyl alcohol, we first need the balanced chemical equation for the reaction. The combustion of ethyl alcohol (C₂H₅OH) typically occurs according to the following equation:
C₂H₅OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 3 H₂O (g)
According to the stoichiometry of the reaction, one mole of ethyl alcohol reacts with three moles of oxygen to produce two moles of carbon dioxide. Given that we have 1.42 moles of oxygen and 0.900 moles of ethyl alcohol, we need to find the limiting reactant, which is the reactant that will run out first and dictates how much product can be formed.
We can calculate the moles of carbon dioxide produced based on the stoichiometry and the limiting reactant. At STP, one mole of a gas occupies 22.414 liters. The reaction tells us that for every 1 mole of ethyl alcohol, 2 moles of carbon dioxide are produced. Since the question provides more moles of oxygen than ethyl alcohol (and the molar ratio is 1:3), ethyl alcohol is the limiting reactant.
Therefore, from 0.900 moles of ethyl alcohol, we will get 1.8 moles of carbon dioxide (0.900 moles × 2). Using the molar volume of a gas at STP (22.414 L/mol), we can calculate the volume of carbon dioxide:
Volume of CO₂ = 1.8 moles × 22.414 L/mol = 40.3452 L
The volume of carbon dioxide produced at STP by the reaction would be approximately 40.3452 liters.