What is the volume of O2 required for the combustion of 1 mole of C5H10 at STP?
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We need to find the volume of oxygen gas required for the complete combustion of one mole of C₅H₁₀ at standard temperature and pressure. This is a stoichiometry problem involving gas volumes.
First, we need to write the balanced chemical equation for the complete combustion of C₅H₁₀. The unbalanced equation shows C₅H₁₀ reacting with oxygen to produce carbon dioxide and water. After balancing, we get: 2 C₅H₁₀ plus 15 O₂ yields 10 CO₂ plus 10 H₂O.
From the balanced equation, we can see that 2 moles of C₅H₁₀ react with 15 moles of O₂. This gives us a mole ratio of 15 moles of oxygen to 2 moles of C₅H₁₀, which simplifies to 7.5 moles of oxygen per mole of C₅H₁₀.
Now we use the molar volume of a gas at STP, which is 22.4 liters per mole. We multiply 7.5 moles of oxygen by 22.4 liters per mole to get 168 liters. Therefore, the volume of oxygen required for the combustion of 1 mole of C₅H₁₀ at STP is 168 liters.
To summarize our solution: We balanced the combustion equation, determined the mole ratio of oxygen to C₅H₁₀, and calculated the volume using the molar volume at STP. The final answer is 168 liters of oxygen gas required for the complete combustion of 1 mole of C₅H₁₀ at standard temperature and pressure.