0.250g of Fe3O4 reacts with excess O2 to give Fe2O3 in 73.2% yield according to the following balanced equation: 4Fe3O4(s)+O2(g)->6Fe2O3(s). What is the theoretical yield of Fe2O3? What is the actual yield of Fe2O3?

Answer 1

To find the theoretical yield of Fe2O3, we first need to calculate the molar mass of Fe3O4 and Fe2O3:

Molar mass of Fe3O4:
Fe = 55.85 g/mol
O = 16.00 g/mol
Fe3O4 = 3(Fe) + 4(O) = 3(55.85) + 4(16.00) = 231.7 g/mol

Molar mass of Fe2O3:
Fe = 55.85 g/mol
O = 16.00 g/mol
Fe2O3 = 2(Fe) + 3(O) = 2(55.85) + 3(16.00) = 159.7 g/mol

Next, we need to calculate the moles of Fe3O4:
0.250g / 231.7 g/mol = 0.00108 mol

According to the balanced equation, 4 moles of Fe3O4 react to produce 6 moles of Fe2O3. Therefore, the theoretical yield of Fe2O3 can be calculated as follows:

0.00108 mol Fe3O4 * (6 mol Fe2O3 / 4 mol Fe3O4) = 0.00162 mol Fe2O3

Finally, we calculate the theoretical yield of Fe2O3 in grams:

0.00162 mol * 159.7 g/mol = 0.259 g Fe2O3

Therefore, the theoretical yield of Fe2O3 is 0.259g.

Next, we need to calculate the actual yield of Fe2O3, which is given as 73.2% of the theoretical yield:

73.2% of 0.259g = 0.1896g

Therefore, the actual yield of Fe2O3 is 0.1896g.