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...

Question

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 522 mol522 mol of octane combusts, what volume of carbon dioxide is produced at 38.0 ∘C38.0 ∘C and 0.995 atm?

Answer 1

Answer #1

Balanced equation is--

2C8H18(l) + 25O2(g) --> 16CO2(g) + 18H2O(l)

Moles of CO2 produced = (16/2)* moles of octane

= 8*522

= 4176 mole

Now, given T = 38°C = 311.15 K

P = 0.995 atm

So, volume = nRT/P

= (4176*0.082*311.15)/0.995

= 107083.133 L

So, volume of CO2 produced = 107083.1 L

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Answer 2

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