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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 KMnOâ‚„ 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:

  • KMnOâ‚„: 2 moles
  • Kâ‚‚MnOâ‚„: 1 mole
  • MnOâ‚‚: 1 mole
  • Oâ‚‚: 1 mole

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

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

moles of KMnOâ‚„= 0.54

The molar mass of KMnOâ‚„ 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 KMnOâ‚„ to produce 0.27 moles of O2, assuming 100% conversion.

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