Question

(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed?

Answer 1

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

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: 122.45 g/mole
• KCl: 74.45 g/mole
• O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
• KCl: 2 moles ×74.45 g/mole= 148.9 grams
• O₂: 3 moles ×32 g/mole= 96 grams

Ideal gas law

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 the gas occupies.
• T is the temperature of the gas.
• R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
• n is the number of moles of the gas.

Number of O₂ produced.

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

• P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
• V= 165 mL= 0.165 L (being 1000 mL= 1 L)
• n= ?
• R= 0.082 $\frac{atmL}{molK}$
• T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 $\frac{atmL}{molK}$× 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 $\frac{atmL}{molK}$× 303 K)

n= 0.0059 moles

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

Mass of KClO₃ required

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

$mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}$

mass of KClO₃= 0.48144 grams

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

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ideal gas law:

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