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8.
Sulfur trìoxide decomposes to sulfur dioxide and oxygen gas as
shown in the following reaction:
2503 (g) 2502 (g) + O2 (g)
Given a 240.0 g sample of sulfur trioxide (MM = 80.1 g/mol), how many
grams of oxygen (MM = 32.0 g/mol) are produced, assuming the
decomposition goes to completion?
g
Your answer should be rounded to three significant figures. Do not include units in your
answer

Sagot :

Ardni313idn

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

CintiaSalazaridn

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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