The combustion of octane, C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) If 498 mol of...

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Question

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?

Answer 1

Answer #1

n = 498 mot of octane. P=0.995 atm T= 14°C = 2.73 + 14 = V = ? R = 0.082057 Loatm-k-moll No. of moles of coz produced, 208 Hi

Answer 2

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