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Taking into account the reaction stoichiometry, 0.041 moles od KCl are formed when 5 grams of KClO₃ is decomposed.

Reaction stoichiometry

In first place, the balanced reaction is:

2 KClO₃ → 2 KCl + 3 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:

  • KClO₃: 2 moles
  • KCl: 2 moles
  • O₂: 3 moles

The molar mass of the compounds is:

  • KClO₃: 122.55 g/mole
  • KCl: 74.55 g/mole
  • O₂: 32  g/mole

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

  • KClO₃: 2 moles ×122.55 g/mole= 245.1 grams
  • KCl: 2 moles ×74.55 g/mole= 149.1 grams
  • O₂: 3 moles ×32  g/mole= 96 grams

Mass of KCl formed

The following rule of three can be applied: if by reaction stoichiometry 245.1 grams of KClO₃ form 2 moles of KCl, 5 grams of KClO₃ form how many moles of KCl?

[tex]moles of KCl=\frac{5 grams of KClO_{3}x2 moles of KCl }{245.1 grams of KClO_{3}}[/tex]

moles of KCl= 0.041 moles

Then, 0.041 moles od KCl are formed when 5 grams of KClO₃ is decomposed.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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