How many grams of O2 are needed to produce 45.8 grams of Fe2O3 in the following reaction? 4Fe(s)+3O2(g) - 2Fe2O3(s)
To solve this problem, we can use the stoichiometry of the reaction to determine the amount of O2 needed to produce a given amount of Fe2O3.
In the balanced chemical equation, we see that the coefficient in front of Fe2O3 is 2. This means that in order to produce 2 moles of Fe2O3, we need 3 moles of O2.
To find the number of moles of Fe2O3, we divide the given mass of Fe2O3 by its molar mass. The molar mass of Fe2O3 can be calculated by adding up the atomic masses of its constituent atoms: (2 x molar mass of Fe) + (3 x molar mass of O).
The molar mass of Fe is approximately 55.85 g/mol, and the molar mass of O is approximately 16.00 g/mol. Therefore, the molar mass of Fe2O3 is:
(2 x 55.85 g/mol) + (3 x 16.00 g/mol) = 159.70 g/mol
Now, we can calculate the number of moles of Fe2O3:
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
moles of Fe2O3 = 45.8 g / 159.70 g/mol ≈ 0.287 mol
Since the stoichiometric ratio between Fe2O3 and O2 is 2:3, we can set up a proportion to find the number of moles of O2 needed:
(0.287 mol Fe2O3) / (2 mol Fe2O3) = (x mol O2) / (3 mol O2)
Cross-multiplying and solving for x, we get:
x = (0.287 mol Fe2O3) * (3 mol O2 / 2 mol Fe2O3)
x = 0.4315 mol O2
Now, to find the mass of O2 needed, we multiply the number of moles of O2 by its molar mass:
mass of O2 = moles of O2 * molar mass of O2
mass of O2 = 0.4315 mol * 32.00 g/mol ≈ 13.8 g
Therefore, approximately 13.8 grams of O2 are needed to produce 45.8 grams of Fe2O3 in the given reaction.