Question

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 418 mol418 mol...

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 418 mol418 mol of octane combusts, what volume of carbon dioxide is produced at 14.0 ∘C14.0 ∘C and 0.995 atm?

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Answer #1

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)

2 moles of C8H18 react with excess of O2 to gives 16 moles of CO2

418 moles of C8H18 react with excess of O2 to gives = 16*418/2 = 3344 moles of CO2

T = 14+273   = 287K

P   = 0.995atm

n = 3344moles

PV = nRT

V    = nRT/P

       = 3344*0.0821*287/0.995

        = 79190L >>>>answer

The volume of CO2 = 79190L >>>>answer

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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 418 mol418 mol...
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