Final answer:
From 86.17 grams of C₆H₁₄ and excess oxygen, approximately 264.054 grams of CO₂ can be produced by first balancing the combustion equation, then using stoichiometry to find the moles of C₆H₁₄, and finally converting those moles to mass of CO₂.
Step-by-step explanation:
The question asks for the mass of carbon dioxide (CO₂) that can be produced from 86.17 grams of C₆H₁₄ (hexane) when reacted with excess oxygen. To answer this, we'll require the balanced chemical equation for the combustion of hexane and then use stoichiometry to calculate the mass of CO₂ produced.
Firstly, let's balance the chemical equation for the combustion of hexane:
2 C₆H₁₄(l) + 19 O₂(g) → 12 CO₂(g) + 14 H₂O(g)
Next, we'll calculate the moles of C₆H₁₄ using its molar mass (86.18 g/mol for C₆H₁₄):
Moles of C₆H₁₄ = Mass / Molar mass = 86.17 g / 86.18 g/mol = ~1 mol C₆H₁₄
According to the balanced equation, the combustion of 2 moles of C₆H₁₄ produces 12 moles of CO₂. Since we have approximately 1 mole of C₆H₁₄, it will produce approximately 6 moles of CO₂.
Finally, we convert the moles of CO₂ to mass:
Mass of CO₂ = Moles of CO₂ x Molar mass of CO₂ = 6 moles x 44.009 g/mol = 264.054 g CO₂
So, 86.17 grams of C₆H₁₄ with excess oxygen will produce approximately 264.054 grams of CO₂.