Answer:
The percentage yield of O2 is 66.7%
Explanation:
Reaction for decomposition of potassium chlorate is:
2KClO₃ → 2KCl + 3O₂
The products are potassium chloride and oxygen.
Let's find out the moles of chlorate.
Mass / Molar mass = Moles
12.3 g / 123 g/mol = 0.1 mol
So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.
Then, 0.1 mol of chlorate may produce (0.1 .3)/ 2 = 0.15 moles
Let's convert the moles of produced oxygen, as to find out the theoretical yield.
0.15 mol . 32 g/ 1mol = 4.8 g
To calculate the percentage yield, the formula is
(Produced Yield / Theoretical yield) . 100 =
(3.2g / 4.8g) . 100 = 66.7 %
The branch of science which deals with the study of chemicals and their bond is called chemistry.
The correct percentage yield of O2 is 66.7%
The Reaction for decomposition of potassium chlorate is as follows:-
[tex]2KClO_3 <---> 2KCl + 3O_2[/tex]
The formula is as follows:- [tex]\frac{Mass }{Molar\ mass} = Moles[/tex]
After putting the value in the question is:-
[tex]\frac{12.3 g}{123} = 0.1 mol[/tex]
So the ratio present in the reaction is 2:3.
Therefore, the 0.1 mole of chlorate produce [tex]\frac{(0.1 *3)}{2} = 0.15 moles[/tex]
Convert them into molar mass is:-
[tex]\frac{0.15*32 g}{1mol} = 4.8 g[/tex]
The percentage of the compound is [tex]\frac{(3.2g}{4.8g}* 100 = 66.7 %[/tex].
Hence, the correct answer is 66.7%
For more information, refer to the link:-
https://brainly.com/question/19524691
