The combustion of octane, C8H18, proceeds according to the reaction shown: 2C8H18 (l) + 25O2 (g) ⟶ 16CO2 (g) + 18H2O (l) If 587 moles of octane combust, what volume of carbon dioxide is produced at 38.0 °C and 0.995 atm?
Added by Rosario P.
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Given that 2 moles of octane produce 16 moles of carbon dioxide, Number of moles of carbon dioxide produced = (16/2) * 587 Number of moles of carbon dioxide produced = 8 * 587 Number of moles of carbon dioxide produced = 4696 moles Show more…
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The combustion of octane, C8H18, proceeds according to the reaction shown.2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)If 426 mol of octane combusts, what volume of carbon dioxide is produced at 32.0 ∘C and 0.995 atm?
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2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g) Calculate the volume of carbon dioxide produced when 1.0805 g of octane, C8H18(l), combusts at STP. The molar mass of C8H18 is 114.26 g/mol. Express your answer with the appropriate significant figures and unit. Options: 1.695 L CO2(g) 18.900 L CO2(g) 1.69 L CO2(g) 3.33 L CO2(g) 3.33 g CO2(g)
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