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?

Answer 1

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.