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How
many grams of oxygen gas
are
produced
when 2.43x10(4) g of KClO3
are
completely
reacted
according
to
the
following
chemical
equation:
2 KClO₃(s)

2 KCl(s)
+
3 O₂(g)

Sagot :

Answer:

2 KCl(s)

Explanation:

just took the exam

Taking into account the reaction stoichiometry, 9717.75 grams of oxygen gas  are  produced  when 2.43×10⁴ grams of KClO₃  are  completely  reacted.

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 each compound 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

Then you can apply the following rule of three: if by reaction stoichiometry 245.1 grams of KClO₃ form 96 grams of O₂, 2.43×10⁴ grams of KClO₃ form how much mass of O₂?

[tex]mass of O_{2} =\frac{2.43x10^{4} grams of KClO_{3} x 96 grams of O_{2}}{245.1grams of KClO_{3} }[/tex]

mass of O₂= 9517.75 grams

In summary, 9717.75 grams of oxygen gas  are  produced  when 2.43×10⁴ grams of KClO₃  are  completely  reacted.

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