Octane (gasoline) undergoes combustion in automobile engine. The balanced equati

Question

Octane (gasoline) undergoes combustion in automobile engine. The balanced equation for this reaction is C8H18(l) + 12.5O2(g) --> 8CO2(g) + 9H2O(g)
What is the volume at 273 K and 1.00 atm of the oxygenrequired for the complete combustion of 4.0 moles of octane? a. 112 L b. 1.12 x 10e3 L c. 560 L d. 5.6 x10e3 L
The answer is B but how? Octane (gasoline) undergoes combustion in automobile engine. The balanced equation for this reaction is C8H18(l) + 12.5O2(g) --> 8CO2(g) + 9H2O(g)
What is the volume at 273 K and 1.00 atm of the oxygenrequired for the complete combustion of 4.0 moles of octane? a. 112 L b. 1.12 x 10e3 L c. 560 L d. 5.6 x10e3 L
The answer is B but how?

Explanation / Answer

from the above reaction        1 mol of Octane requires 12.5mols of Oxygen now 4 moles of Octane requires 4*12.5 mols of Oxygen                                                   50 moles       we know that PV = nRT     given data           P = 1atm            V =?            n = 50moles           R = 0.0821L.atm/K.mol           T = 273K     therefore V = nRT/P                       = (50 mole)(0.0821 L.atm/mol.K)(273K)/1 atm                       = 1120.66 L                       = 1.12*103       hence option b is correct

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