A 2.00g sample of Fe3O4 is reacted with with 8.00g of O2 to produce Fe2O3.
a) What mass of Fe2O3 will be produced?
b) What mass of Fe2O3 will actually be produced if the percentage yield for reaction is 85% produced?
Start with writing a balanced equation:
4Fe3O4+O2 => 6Fe2O3
Then use stoichiometric coefficients and molar masses to solve the remainder of the problem.
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To answer these questions, we need to first determine the balanced chemical equation for the reaction. The reaction between Fe3O4 and O2 produces Fe2O3.
The balanced chemical equation is:
Fe3O4 + O2 -> Fe2O3
Now let's calculate the masses of Fe2O3 that will be produced.
a) To determine the mass of Fe2O3 produced, we need to use stoichiometry to find the ratio between Fe3O4 and Fe2O3 in the balanced equation.
The molar mass of Fe3O4 is:
Fe: 55.85 g/mol x 3 = 167.55 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total: 167.55 g/mol + 64.00 g/mol = 231.55 g/mol
Now we can calculate the number of moles of Fe3O4:
Number of moles = mass / molar mass
Number of moles = 2.00 g / 231.55 g/mol = 0.00863 mol
From the balanced equation, we can see that the stoichiometric ratio between Fe3O4 and Fe2O3 is 1:1. Therefore, the number of moles of Fe2O3 produced will also be 0.00863 mol.
Finally, we can find the mass of Fe2O3 produced using its molar mass:
Mass = number of moles x molar mass
Mass = 0.00863 mol x (Fe: 55.85 g/mol x 2 + O: 16.00 g/mol x 3) = 0.950 g
Therefore, the mass of Fe2O3 produced is 0.950 g.
b) To calculate the mass of Fe2O3 actually produced if the percentage yield is 85%, we need to multiply the calculated mass by the percentage yield.
Mass of Fe2O3 actually produced = Mass of Fe2O3 produced x Percentage yield / 100
Mass of Fe2O3 actually produced = 0.950 g x 85 / 100 = 0.808 g
Therefore, the mass of Fe2O3 actually produced, considering an 85% yield, is 0.808 g.