Posted by Rafael on Tuesday, September 4, 2012 at 10:01pm.
2C6H6 + 15O2 ==> 12CO2 + 6H2O
6.38 mol C6H6 x (12 mols CO2/2 mols C6H6) = 38.3 mols CO2 formed if we started with 6.38 mol C6H6 and had all of the O2 we needed.
6.03 mol O2 x (12 mols CO2/15 mol O2) = 4.82 mols CO2 formed if we started with 6.03 mol O2 and had all of the C6H6 we needed.
In limiting reagent problems the correct answer is ALWAYS the smaller value; therefore, the limiting reagent is O2 and we have some C6H6 remain unreacted. We will have a maximum of 4.82 mols CO2 formed.
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