Final answer:
The question involves applying concepts such as Avogadro's law, the ideal gas law, and the combined gas law to determine the volume of gases produced from the combustion of ethane (C₂H₆) at non-STP conditions.
Step-by-step explanation:
Combustion of C₂H₆ at STP
The subject question is asking about the volume of gases produced from the combustion of ethane (Â₂Ã₂₆) at non-STP conditions. According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain the same number of particles. At STP (standard temperature and pressure), one mole of any gas occupies 22.4 L. In the balanced chemical reaction for the combustion of ethane:
Â₂Ã₂₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (g)
You can see that the combustion of 2 moles of ethane produces 4 moles of carbon dioxide (À₂) and 6 moles of water (H₂Â₂Å₂). To find the volumes of CO₂ and H₂O produced from the combustion of 1 L of C₂Ã₂₆ at STP, the ideal gas law and molar volume will be used.
The initial volume of ethane, assuming it behaves as an ideal gas at STP, represents 1/22.4 of a mole. Since the reaction shows a 2:4 molar ratio between ethane and carbon dioxide, and 2:6 ratio with water vapor, you would multiply this fraction by 4 and by 6 to find the volume of CO₂ and H₂O, respectively, at STP.
Given the non-STP conditions at 565°C and 0.872 atm, the combined gas law will be applied to convert the volumes from STP to the desired conditions. The final volume for each gas can be added together to get the total volume of products.
The partial pressure of each product can be calculated by using Dalton's Law, since the final mixture is at the same temperature and pressure.