4 Fe + 3 O2 -> 2 Fe2O3
I don't agree with your answers for theoretical yield
Part 2. (actual yield/theor yield)*100 = ?
Part 3. Same equation as for #2 but plug in 95% for yield and calculate yield.
Post your work if you get stuck.
actual yield
4 Fe + 3 O2 -> 2 Fe2O3
1. If I react 3.00g Fe and 5.75g O2, what is the theoretical yield?
2. If I’m only able to produce 3.5g, what is my percent yield?
3. If I try again, and get a 95.0% yield, what is my actual yield?
For the theo yield I got 0.026857655 mol of Fe, I need help on where to continue
2 answers
%Yield = [Actual Lab Yield mass/Theoretical Calculated Yield mass]100%
Theoretical Yield calculations are based upon the limited reactant in the problem; that is, in the posted problem the limiting reactant is the iron (Fe) with Oxygen remaining in excess giving 0.027 mole Fe2O3 (Theoretical). For the grams multiply by the formula weight of Fe2O3 (160 g/mole) => 4.32 gram theoretical yield.
%Yield = (Lab Yield/Theoretical Yield)100% =(3.5-g/4.32-g)100% = 81%
For 95% yield (Trial 2) => mass yield = 95% of Theoretical Yield = 0.95(4.32) grams = 4.10-grams (Trial 2).
Theoretical Yield calculations are based upon the limited reactant in the problem; that is, in the posted problem the limiting reactant is the iron (Fe) with Oxygen remaining in excess giving 0.027 mole Fe2O3 (Theoretical). For the grams multiply by the formula weight of Fe2O3 (160 g/mole) => 4.32 gram theoretical yield.
%Yield = (Lab Yield/Theoretical Yield)100% =(3.5-g/4.32-g)100% = 81%
For 95% yield (Trial 2) => mass yield = 95% of Theoretical Yield = 0.95(4.32) grams = 4.10-grams (Trial 2).
1 answer