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Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times, it can be easier to chemically prepare occasionally used gases. For example, oxygen gas can be prepared by heating KMnO4(s) according to the following chemical reaction:
2KMnO4(s) → K2MnO4(s) + MnO2(s) + O2(g)
How many grams of KMnO4 would you need to produce 0.27 moles of O2, assuming 100% conversion?

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CintiaSalazar

Answer:

You need 85.32 grams of KMnO₄ to produce 0.27 moles of O2, assuming 100% conversion.

Explanation:

The balanced chemical reaction is:

2 KMnO₄ (s) → K₂MnO₄ (s) + MnO₂ (s) + O₂ (g)

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

  • KMnO₄: 2 moles
  • K₂MnO₄: 1 mole
  • MnO₂: 1 mole
  • O₂: 1 mole

Then you can apply the following rule of three: if by stoichiometry 1 mole of O₂ is produced by 2 moles of KMnO₄, 0.27 moles of O₂ are produced by how many moles of KMnO₄?

[tex]moles of KMnO_{4} =\frac{0.27 moles of O_{2} *2moles of KMnO_{4} }{1mole of O_{2} }[/tex]

moles of KMnO₄= 0.54

The molar mass of KMnO₄ is 158 [tex]\frac{g}{mol}[/tex].

Then the amount of mass present in 0.54 moles of the compound can be calculated by:

0.54 moles* 158.034 [tex]\frac{g}{mol}[/tex]= 85.32 grams

You need 85.32 grams of KMnO₄ to produce 0.27 moles of O2, assuming 100% conversion.

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