Posted by Lisa on Monday, March 26, 2012 at 9:21pm.
I work limiting reagent problems just like I do regular stoichiometry problems, except that I work each reactant separately.
First Fe and we assume we have all of the O2 needed. We use the coefficients in the balanced equation to make the conversions.
0.5 mol Fe x (2 mol Fe2O3/4 mol Fe) = 0.5 x 1/2 = 0.25 mol Fe2O3 produced.
Next O2 and we assume we have all of the Fe needed.
0.4 mol O2 x (2 mol Fe2O3/3 mol O2) = 0.4 x 2/3 = 0.267 mol Fe2O3 produced.
Obviously both answers (0.25 and 0.267)can't be correct; the correct one in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, Fe is the limiting reagent and 0.25 mol Fe2O3 will be produced. What about the oxygen? The needed amount will react and leave the remainder unreacted. We can even calculate how much reacts and how much is left.
0.5 mol Fe x (3 mol O2/4 mol Fe) = 0.5 x 3/4 = 0.375 mol O2 used in the reaction and 0.4 mol - 0.375 mol = 0.025 mol O2 remains unreacted.
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