Final answer:
To solve this problem, we need to use stoichiometry and the given unbalanced equation to determine the number of grams of Fe₂O₃ produced when FeS₂ reacts with O₂. By balancing the equation, converting the given masses to moles, and using the stoichiometric ratio, we can calculate that 66.7 grams of Fe₂O₃ will be produced.
Step-by-step explanation:
To solve this problem, we need to use stoichiometry and the given unbalanced equation.
First, we balance the equation by placing coefficients in front of each compound to make sure the number of atoms on each side is equal.
The balanced equation would be:
FeS₂ + 4O₂ → Fe₂O₃ + 2SO₂
Now, we can use the balanced equation to set up a stoichiometric ratio. The molar mass of Fe₂O₃ is 159.7 g/mol. We have 36.8 g of FeS₂ and 22.0 g of O₂. We calculate the number of moles for each substance by dividing the mass by their respective molar masses.
For FeS₂: 36.8 g / (87.92 g/mol) = 0.419 mol
For O₂: 22.0 g / (32.0 g/mol) = 0.688 mol
From the balanced equation, we can see that the stoichiometric ratio between FeS₂ and Fe₂O₃ is 1:1.
Therefore, 0.419 mol of FeS₂ will react to form 0.419 mol of Fe₂O₃.
To calculate the mass of Fe₂O₃, we multiply the number of moles by its molar mass:
0.419 mol × 159.7 g/mol = 66.7 g
Therefore, 66.7 grams of Fe₂O₃ will be produced when 36.8 grams of FeS₂ reacts with 22.0 grams of O₂.