Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.
<h3>Reaction stoichiometry</h3>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:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
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:
165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:
Replacing in the ideal gas law:
0.888231 atm× 0.165 L = n× 0.082 × 303 K
Solving:
n= (0.888231 atm× 0.165 L)÷ (0.082 × 303 K)
<u><em>n= 0.0059 moles</em></u>
Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.
<h3>Mass of KClO₃ required</h3>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?
<u><em>mass of KClO₃= 0.48144 grams</em></u>
Finally, 0.48144 grams of KClO₃ was decomposed.
Learn more about
the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
ideal gas law:
Answer:
Explanation:
The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"
The half-life of the element is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows: