contestada

Determine the volume of O2 (at STP) formed when 50.0 g of KClO3 decomposes according to the following reaction. The molar mass for KClO3 is 122.55 g/mol. 22KClO3 (s) --> 2 KCl(s) + 3 O2(g)

Respuesta :

CintiaSalazar

Taking into account the reaction stoichiometry, ideal gas law and the definition of STP conditions, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

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 for KClO₃ is 122.55 g/mol. Then, by reaction stoichiometry, the following mass quantity of the compound participates in the reaction:

[tex]2 molesx\frac{122.55 g}{1 mole} =[/tex]245.1 grams of KClO₃

 

Then it is possible to apply the following rule of three: if by reaction stoichiometry 245.1 grams of KClO₃ form 3 moles of O₂, 50 grams of KClO₃ form how many moles of O₂?

[tex]amount of moles of O_{2} =\frac{50 grams of KClO_{3} x3 moles of O_{2} }{245.1grams of KClO_{3}}[/tex]

amount of moles of O₂= 0.612 moles

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Then, in this case you know:

  • P= 1 atm
  • V= ?
  • n= 0.612 moles
  • R= 0.082 [tex]\frac{atmL}{molK}[/tex]
  • T= 0 C= 273 K

Replacing in the ideal gas:

1 atm× V= 0.612 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 273 K

Solving:

[tex]V=\frac{0.612 molesx 0.082\frac{atmL}{molK} x 273 K}{1 atm}[/tex]

V=13.70 L

Finally, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

Learn more:

  • https://brainly.com/question/2253160?referrer=searchResults
  • https://brainly.com/question/14709513?referrer=searchResults
  • https://brainly.com/question/20158093?referrer=searchResults
  • https://brainly.com/question/6240287?referrer=searchResults
  • https://brainly.com/question/10307303?referrer=searchResults

STP conditions refer to the standard temperature and pressure. The volume of the oxygen at STP will be 13.70 L when 50 grams of potassium chlorate decomposes.

The chemical equation of decomposition if potassium chlorate can be written as:

[tex]\text{KClO}_3\;\rightarrow\; 2\;\text{KCl + 3\; O}_2[/tex]

From the stoichiometry of the reaction, it can be observed as:

  • 2 moles of potassium chlorate
  • 2 moles of potassium chloride
  • 3 moles of oxygen

Now, calculating the mass of potassium chlorate participating in the reaction:

  • [tex]2\;\text{moles}\;\times\dfrac{122.5\;\text g}{1 \;\text{mole}} &= 245.1 \;\text {g}\;\text {KClO}_3[/tex]

Now, substituting the values to calculate the moles of oxygen:

[tex]\text{amount of moles oxygen}&=\dfrac{50\;\text g} {245.1 \;\text g} \times 3\;\text{moles of oxygen}[/tex]

  • Moles of oxygen = 0.612 moles

From the ideal gas equation:

  • PV = nRT, where,

P is pressure, Tis  the temperature, and the volume, V. R is ideal gas constant.

Substituting the values:

  • 1 atm x V = 0.612 moles x 0.082 x 273 K
  • V = 13.70 L

Therefore, the 13.70 L of oxygen at STP will be formed when 50 grams of potassium chlorate decomposes into potassium and oxygen.

To know more about ideal gas equation, refer to the following link:

https://brainly.com/question/1151028?referrer=searchResults

Otras preguntas