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If you begin with 5.000 grams of KClO3(s) how many moles and KClO3(s)
will be used, and b) how many grams, moles, and molecules of the product species will
be formed?

If You Begin With 5000 Grams Of KClO3s How Many Moles And KClO3s Will Be Used And B How Many Grams Moles And Molecules Of The Product Species Will Be Formed class=

Sagot :

mickymike92

Based on the given data, the amount of products from 5.00 g of KClO₃ is:

  • 0.04 moles of KCl
  • 2.98 g KCl
  • 2.41 * 10²² molecules of KCl
  • 0.06 moles of O₂.
  • 1.92 g of O₂
  • 3.61 * 10²² molecules of O₂

What amount of products is formed from the decomposition of KClO₃?

The decomposition of KClO₃ is given by the equation below:

2 KClO₃  -----> 2 KCl + 3 O₂

2 moles of KClO₃ produces 2 moles of KCl and 3 moles of O₂

Molar mass of KClO₃ is 122.5 g/mol

molar mass of KCl = 74.5 g

molar mass of O₂ = 32 g

Moles of KClO₃ in 5.00 g = 5.00/122.5

  • moles of KClO₃ = 0.04 moles

For KCl

a. moles: 0.04 moles of KClO₃ will produce 0.04 moles of KCl

b. mass of KCl = 0.04 * 74.5 = 2.98 g KCl

c. number of molecules of KCl = 0.04 * 6.02 * 10²³ = 2.41 * 10²² molecules of KCl

For O₂:

a. moles of O₂; 0.04 moles of KClO₃ will produce = 0.06 moles of O₂.

b. mass of O₂ = 0.06 * 32 g = 1.92 g of O₂

c. number of molecules: 0.06 * 6.02 * 10²³ = 3.61 * 10²² molecules of O₂

Therefore, the amount of products from 5.00 g of KClO₃ is:

  • 0.04 moles of KCl
  • 2.98 g KCl
  • 2.41 * 10²² molecules of KCl
  • 0.06 moles of O₂.
  • 1.92 g of O₂
  • 3.61 * 10²² molecules of O₂

Learn more about moles and molecules at: https://brainly.com/question/26135244

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