Respuesta :
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.
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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.
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