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If 36.8 grams of FeS₂ is allowed to react with 22.0 grams of O₂ according to the following unbalanced equation, how many grams of Fe₂O₃ are produced?

FeS₂+O₂→Fe₂O₃+SO₂

User LacOniC
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1 Answer

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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.

User Darkdeamon
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