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 4.10×102 mol4.10×102 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 4.10×102 mol4.10×102 mol of octane combusts, what volume of carbon dioxide is produced at 12.0 ∘C12.0 ∘C and 0.995 atm?

science chemistry

Answer 1

Answer #1

2 Cetig + 250 - 1600 + 184,0 mole of octane = mole of co, = 4.10x lot = 410 mole 410x8= 3280 mole volume of co = nx RXT = 328

Answer 2

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