Calculate the theoretical yield of iron(lll) oxide (Fe2O3)

Question

Iron reacts with oxygen at high temperatures to form Iron(lll) oxide.

Suppose 24.3g of iron (Fe) is reacted with 29.1 g of oxygen (O2)

4Fe(s)+3O2(G) - 2Fe2O3(S)

Select the limiting reactant

the reaction produces 6.66g of Fe2O3. What is the percent yield of the reaction

Answer 1

Using a periodic table, Iron (Fe) has an atomic mass of 55.84 g/mol. Since we start with 24.3 g of Fe, we have (24.3 g)/(55.84 g/mol) = 0.435 mol Fe. Since there are 4 moles of Fe for every mole of reaction, we have (0.435 mol Fe) / (4 mol Fe/mol RXN) = 0.109 mol RXN.

Oxygen (O) has an atomic mass of 15.999 g/mol. The reactant with oxygen is O₂, so the molar mass of this compound is 15.999 g/mol * (2 mol O) / (1 mol O₂) = 31.998 g/mol. Since we start with 29.1 g of O, we have (29.1 g)/(31.998 g/mol) = 0.909 mol O. Since there are 3 moles of O₂ for every mole of reaction, we have (0.909 mol O₂) / (3 mol O₂ /mol RXN) = 0.303 mol RXN.

The starting amount of Fe allows for 0.109 mol RXN and the starting amount of O₂ allows for 0.303 mol RXN, so Fe will run out first. Therefore, Fe is the limiting reactant of this reaction.

0.109 mol RXN will happen with these starting amounts. For every mole of reaction, the products created are 2 moles of Fe₂O₃. So, there are (0.109 mol RXN) * (2 mol Fe₂O₃ /1 mol RXN) = 0.218 mol Fe₂O₃.

The molar mass of Fe₂O₃ is 2 * (55.84 g/mol) + 3 * (15.999 g/mol) = 159.7 g/mol Fe₂O₃.

So, there are 159.7 g/mol Fe₂O₃ * 0.218 mol Fe₂O₃ = 34.7 g Fe₂O₃.

6.66g Fe₂O₃ is produced, when it should be 34.7g Fe₂O₃.

So, the percent yield of the reaction is 6.66g/34.7g * 100 = 19.2% Yield.