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Sagot :
Mass of Oxygen : 47.9
Further explanation
Given
2S03 (g)⇒ 2S02 (g) + O2 (g)
240 g SO3
Required
mass of Oxygen
Solution
mol SO3 :
= 240 g : 80.1 g/mol
= 2.996
From the equation, mol O2 :
= 1/2 x mol SO3
= 1/2 x 2.996
= 1.498
mass O2 :
= 1.498 mol x 32 g/mol
= 47.9
Taking into account the reaction stoichiometry, given a 240.0 g sample of sulfur trioxide, 47.95 grams of oxygen are produced, assuming the decomposition goes to completion.
Reaction stoichiometry
In first place, the balanced reaction is:
2 SO₃ → 2 SO₂ + O₂
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:
- SO₃: 2 moles
- SO₂: 2 moles
- O₂: 1 mole
The molar mass of the compounds is:
- SO₃: 80.1 g/mole
- SO₂: 64.1 g/mole
- O₂: 32 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- SO₃: 2 moles ×80.1 g/mole= 160.2 grams
- SO₂: 2 moles ×64.1 g/mole= 128.2 grams
- O₂: 1 mole ×32 g/mole= 32 grams
Mass of oxygen produced
The following rule of three can be applied: if by reaction stoichiometry 160.2 grams of SO₃ form 32 grams of O₂, 240 grams of SO₃ form how much mass of O₂?
[tex]mass of O_{2} =\frac{240 grams of SO_{3}x 32 grams of O_{2} }{160.2grams of SO_{3}}[/tex]
mass of O₂= 47.94 grams
Given a 240.0 g sample of sulfur trioxide, 47.95 grams of oxygen are produced, assuming the decomposition goes to completion.
Learn more about the reaction stoichiometry:
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