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If the percent yield for the following reaction is 65.0%, how many grams of KClO3 are needed to produce 8.00 g of O2?

Sagot :

Taking into account definition of percent yield, 31.42 grams of KClO₃ are needed to produce 8.00 g of O₂.

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

Percent yield

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield= (actual yield÷ theorical yield)×100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Percent yield for the reaction in this case

In this case, you know:

  • actual yield= 8 grams
  • theorical yield= ?
  • percent yield= 65%

Replacing in the definition of percent yields:

65= (8 grams÷ theorical yield)×100

Solving:

theorical yield= (8 grams÷ 65)×100

theorical yield= 12.31 grams

Then, this means that you need to find the mass of KClO₃ would theoretically produce 12.31 g of O₂.

Mass of KClO₃ required

The following rule of three can be applied: If by reaction stoichiometry 96 grams of O₂ are produced by 245.1 grams of KClO₃, 12.31 grams of O₂ are produced by how much mass of KClO₃?

[tex]mass of KClO_{3} =\frac{12.31 grams of O_{2}x245.1 gramsof KClO_{3} }{96 grams of O_{2}}[/tex]

mass of KClO₃= 31.43 grams

Finally, 31.42 grams of KClO₃ are needed to produce 8.00 g of O₂.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

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