If 8.65 grams of iron (iii) oxide react with 7.85 grams of carbon monoxide to produce 5.25 g of pure iron, what are the theoretical yield and percent yield of this reaction? be sure to show the work that you did to solve this problem. unbalanced equation: fe2o3 + co yields fe + co2

1) Balanced equation

Fe2O3 + 3CO ---> 2Fe + 3CO2

2) Theoretical molar ratios

1 mol Fe2O3 : 3 mol CO : 2 mol Fe : 3 mol CO2

3) Covert 8.65 g of Fe2O3 into moles

number of moles = mass in grams / molar mass

molar mass of Fe2O3: 159.69 g/mol

number of moles = 8.65 g / 159.69 g/mol = 0.0542 g/mol Fe2O3

4) Convert 7.85 g of CO to moles

number of moles = mass in grams / molar mass

molar mass of CO = 28.01 g/mol

number of moles = 7.85 g / 28.01 g/mol = 0.280

5) Calculate the limitant reactant

1 mol Fe2O3 / 3 mol CO vs 0.0542 mol Fe2O3 / 0.280

0.33 > 0.19 => CO is in excess and Fe2O3 is the limitant reactant.

6) Theoretical yield

1 mol Fe2O3 / 2 mol Fe = 0.0542 mol Fe2O3 / x

=> x = 0.0542 mol fe2O3 * 2 mol Fe / 1 mol Fe2O3 = 0.1084 mol Fe

convert 0.1084 moles into grams

grams = number of moles * atomic mass = 0.1084 mol * 55.845 g = 6.05 g

Theoretical yield = 6.05 g

7) Percent yield

Percent yield = (actual yield / theoretical yield)*100 = (5.25 / 6.05) * 100 = 86.8%

Percent yield = 86.8%

B.) falsehope this !