3 Fe2O3 + 1 CO → 2 Fe3O4 + 1 CO2
how many grams of Fe2O3 are needed to prepare 255 g of Fe3O4?
a. 118 g
b. 552 g
c. 264 g
d. 383 g
Convert 255 g Fe3O4 to mols. mols = grams/molar mass.
Using the balanced equation, convert mols Fe3O4 to mols Fe2O3.
Convert mols Fe2O3 to grams.
To find out how many grams of Fe2O3 are needed to prepare 255 g of Fe3O4, we need to use stoichiometry, which is the quantitative relationship between the reactants and products in a chemical reaction.
First, we need to determine the molar mass of the substances involved:
- The molar mass of Fe2O3 (iron(III) oxide) is calculated as follows:
Fe = 55.845 g/mol
O = 16.00 g/mol (3 oxygen atoms)
Adding these together, we get:
Fe2O3 = 2(55.845) + 3(16.00) = 159.69 g/mol
- The molar mass of Fe3O4 (iron(II,III) oxide) is calculated as follows:
Fe = 55.845 g/mol (3 iron atoms)
O = 16.00 g/mol (4 oxygen atoms)
Adding these together, we get:
Fe3O4 = 3(55.845) + 4(16.00) = 231.52 g/mol
Now we can set up a ratio based on the balanced chemical equation and use it to calculate the amount of Fe2O3 needed:
3 mol Fe2O3 → 2 mol Fe3O4
255 g Fe3O4 × (1 mol Fe3O4 / 231.52 g Fe3O4) × (3 mol Fe2O3 / 2 mol Fe3O4) × (159.69 g Fe2O3 / 1 mol Fe2O3) = 258.84 g Fe2O3
So, approximately 258.84 g of Fe2O3 are needed to prepare 255 g of Fe3O4.
Therefore, the correct answer is not given.