Before starting the calculation we need to know the density of Octane (C8H18) which is 0.703 g/mL.
Now start by converting the volume of octane from gallons to ml.
(15.0 gallons) x (100ml/0.0264gallons) = 56818.18182 ml of octane
Now find the grams of octane by dividing the volume of octane by the density.
(56818.18182 ml) x (0.703g/1ml) = 39943.18182 grams of octane
Now convert grams of octane to moles by divding by the gram molar mass, 114.23 g/mole.
(39943.18182 grams) / (114.23 g /mol) = 349.6733066 moles of octane
Now use the reaction ration (25 moles O2/2 mole of C8H18) from the balanced equaiton to convert to moles of O2.
(349.6733066 mol C8H18) x (25 moles O2/2 mole C8H18) = 4370.916 moles of O2 are needed.
Last step convert the moles of O2 needed to complete the reation to grams by multiplying by the gram molar mass of O2, 32 g/mole.
(4370.916 mole O2) x (32 g/mole) = 139869.32 grams of O2 are needed to complete the reaction
The combustion of octane, C8H18,C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) If 522 mol522 mol of octane combusts, what volume of carbon dioxide is produced at 38.0 ∘C38.0 ∘C and 0.995 atm?
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The combustion of octane, C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) If 442 mol of octane combusts, what volume of carbon dioxide is produced at 19.0 ∘C and 0.995 atm? ?=
The combustion of octane, C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) If 498 mol of octane combusts, what volume of carbon dioxide is produced at 14.0 ∘C and 0.995 atm?
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