Posted by Anonymous on Tuesday, September 4, 2012 at 7:29pm.
2S + 3O2 -> 2SO3
This example should help you.
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
2S + 3O2 ==> 2SO3
7.0 mols S x (2 mol SO3/2 mol S) = 7.0 mols SO2 formed if we had 7.0 mols S and all the O2 needed.
10 mols O2 x (2 mol SO3/3 mol O2) = 6.67 mols SO2 formed if we had 10 mols O2 and all the S needed.
In limiting reagent questions the smaller value is ALWAYS the correct one to choose; therefore, the limiting reagent is O2 and some S will be remain unreacted.
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