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A sample of oxygen gas is collected over water. The volume of gas
collected is 0.250 L at 26°C and 765 torr total pressure. (Water vapor
pressure at 26°C = 25.22 torr)

How many grams of oxygen gas are collected?

If the oxygen gas collected is the product of decomposition of
potassium chlorate, how many grams of potassium chlorate is
decomposed (assuming 100% yield)?

answer: 0.317g, 0.810g

how do I set the problem up?

Respuesta :

Eduard22sly

A. The mass of the oxygen gas collected is 0.317 g

B. The mass of potassium chlorate that decomposed is 0.81 g

A. How to determine the mass of oxygen collected

  • Volume (V) = 0.250 L
  • Temperature (T) = 26 ˚C = 26 + 273 = 299 K
  • Pressure (P) = 765 – 25.22 = 739.78 torr = 739.78 / 760 = 0.973 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Number of mole (n) =?

The number of mole can be obtained as by using the ideal gas equation as illustrated below:

n = PV / RT

n = (0.973 × 0.25) / (0.0821 × 299)

n = 0.0099 mole

Thus the mass of oxygen collected can be obtained as follow:

  • Mole of oxygen = 0.0099 mole
  • Molar mass of oxygen = 32 g/mol
  • Mass of oxygen =?

Mass of oxygen = mole × molar mass

Mass of oxygen = 0.0099 × 32

Mass of oxygen = 0.317 g

B. How to determine the mass KClO₃

Balanced equation

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

3 moles of O₂ were obtained from 2 moles of KClO₃

Therefore

0.0099 mole of O₂ will be obtained from = (0.0099 × 2) / 3 = 0.0066 moles of KClO₃

Thus, the mass of KClO₃ that decomposed can be obtained as follow:

  • Mole of KClO₃ = 0.0066 mole
  • Molar mass of KClO₃ = 122.5 g/mol
  • Mass of KClO₃ =?

Mass = mole × molar mass

Mass of KClO₃ = 0.0066 × 122.5

Mass of KClO₃ = 0.81 g

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