Question

An unbalanced equation is shown. In this reaction, 200.0 g of FeS2 is burned in 100.0 g of oxygen, and 55.00 g of Fe2O3 is produced.


4FeS2 + 11O2 mc021-1.jpg Fe2O3 + SO2


What is the percent yield of Fe2O3?

Answer 1

Balance Equation
1.) 4FeS2+11O2-2Fe2O3+8SO2

Convert to Moles
2.) 200.0gFe2O3
110.9750 FeS2/Mole
= 1.66701Mole FeS2

100.0g O2
31.99886O2/Mole
=3.12511Mole O2

Convert to Grams
3.) 12511 O2Moles * 2Mole Fe2O3 *159.6882g
=90.735 Grams Fe2O3

Find Excessive/Limiting Reagent
4.) 3.12511 Mole O2 4FeS2
11 O2
= 1.13640Mole FeS20, since more FeS2, then FeS2-excess, O2-Limiting

5.) Get Percent Yield
55.0g Fe2O3
90.735g Fe2O3
=0.6062

0.6062*100= 61% yield