0.300 g 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).
Calculate the theoretical and actual yield of Fe2O3.
First, calculate the theoretical yield of Fe2O3:
1. Calculate the molar mass of Fe3O4:
Fe: 55.845 g/mol
O: 16.00 g/mol
Fe3O4: 3(55.845) + 4(16.00) = 231.532 g/mol
2. Calculate the moles of Fe3O4:
0.300 g / 231.532 g/mol = 0.001297 mol
3. According to the balanced equation, 4 moles of Fe3O4 produce 6 moles of Fe2O3. Therefore, 0.001297 mol of Fe3O4 will produce:
0.001297 mol * (6 mol Fe2O3 / 4 mol Fe3O4) = 0.0019465 mol Fe2O3
4. Calculate the theoretical yield of Fe2O3:
The molar mass of Fe2O3 is 159.688 g/mol
0.0019465 mol * 159.688 g/mol = 0.3108 g Fe2O3
Next, calculate the actual yield of Fe2O3:
The actual yield is given as 61.1% of the theoretical yield.
0.611 * 0.3108 g = 0.1898 g Fe2O3
Therefore, the theoretical yield of Fe2O3 is 0.3108 g and the actual yield is 0.1898 g.