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?
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.