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Taking into account the reaction stoichiometry, 90.33 grams of SO₂ are formed when 124 grams of iron(II) sulfide are used.

Reaction stoichiometry

In first place, the balanced reaction is:

4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • FeS: 4 moles
  • O₂: 7 moles
  • Fe₂O₃: 2 moles
  • SO₂: 4 moles

The molar mass of the compounds is:

  • FeS: 87.85 g/mole
  • O₂: 32 g/mole
  • Fe₂O₃: 159.7 g/mole
  • SO₂: 64 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • FeS: 4 moles ×87.85 g/mole= 351.4 grams
  • O₂: 7 moles ×32 g/mole= 224 grams
  • Fe₂O₃: 2 moles ×159.7 g/mole= 319.4 grams
  • SO₂: 4 moles ×64 g/mole= 256 grams

Mass of sulfur dioxide formed

The following rule of three can be applied: if by reaction stoichiometry 351.4 grams of FeS form 256 grams of SO₂, 124 grams of FeS form how much mass of SO₂?

[tex]mass of SO_{2} =\frac{124 grams of FeSx256 grams ofSO_{2}}{351.4 grams of FeS}[/tex]

mass of SO₂= 90.33 grams

Then, 90.33 grams of SO₂ are formed when 124 grams of iron(II) sulfide are used.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

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