If 5gm KClO3 is decomposed to form KCl and O2. What is the percentage yield of reaction if 1.25 gm of O2 is produced??
To calculate the percentage yield of a reaction, you need to know the theoretical yield and the actual yield. The theoretical yield is the amount of product that would be obtained if the reaction went to completion, while the actual yield is the amount of product actually obtained in the experiment.
In this case, we need to determine the theoretical yield of O2 produced when 5g of KClO3 decomposes. The molar mass of KClO3 is calculated by adding up the atomic masses of potassium (K), chlorine (Cl), and oxygen (O).
The atomic masses are:
- K: 39.10 g/mol
- Cl: 35.45 g/mol
- O: 16.00 g/mol
So, the molar mass of KClO3 is:
(1 * 39.10) + (1 * 35.45) + (3 * 16.00) = 122.55 g/mol
Next, we can calculate the number of moles of KClO3 in 5g by using the molar mass:
Number of moles = Mass / Molar mass = 5g / 122.55 g/mol
Now we need to use the balanced chemical equation for the decomposition of KClO3 to determine the stoichiometry of the reaction. As per the equation, 2 moles of KClO3 decompose to form 3 moles of O2.
So, the number of moles of O2 produced can be calculated by multiplying the number of moles of KClO3 by the stoichiometric ratio:
Number of moles of O2 = (Number of moles of KClO3) * (3 moles of O2 / 2 moles of KClO3)
Now we can calculate the theoretical yield of O2 in grams by using the molar mass of O2:
Theoretical yield = Number of moles of O2 * Molar mass of O2 = (Number of moles of O2) * (32.00 g/mol)
Once you have the theoretical yield, you can calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) * 100
In this case, the actual yield of O2 is given as 1.25g. You can substitute the values into the formula to find the percentage yield of the reaction.