0.300g of Fe3O4 reacts with excess O2 to give Fe2O3 in 61.1% 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 Fe2O3:
Fe: 55.85 g/mol
O: 16.00 g/mol
Fe2O3: (2 x 55.85) + (3 x 16.00) = 159.70 g/mol
Now we can calculate the theoretical yield of Fe2O3:
0.300g Fe3O4 x (6 mol Fe2O3 / 4 mol Fe3O4) x (159.70 g Fe2O3 / 1 mol Fe2O3) = 0.718g Fe2O3
Therefore, the theoretical yield of Fe2O3 is 0.718g.
The actual yield of Fe2O3 is given as 61.1% of the theoretical yield:
0.611 x 0.718g = 0.439 g Fe2O3
Therefore, the actual yield of Fe2O3 is 0.439g.