A 1.24 gram sample of potassium chlorate was heated with manganese dioxide until no addition reaction was visible. THe potassium chloride residue was found to be .72 grams. What was the mass of the oxygen evolved in the experiment?
If there is a balanced equation, if you could please tell me what the products are, thanks :)
The next part: what is the percent, by mass, of oxygen based on the data in the question above?
Thanks for your help, show all work if possible! :)
The reaction is as followed:
2KClO3 ---(MnO2)---> 2KCl + 302
MnO2 is a catalyst
2 moles of KCl produces 3 moles of O2
0.72 g of KCl*(1 mole/74.5513 g)= moles of KCl
moles of KCl*(3 moles of O2/1 mole of KCl)= moles of O2
moles of O2*(31.999g/mole)= mass of O2
(Mass of O2/0.72g +Mass of O2)*100=% of O2 by mass
Isn't it 2 moles of KCL, from the balanced equation? for the 2nd step (mole ratio)
Yes, a typo.
Haha ok, thanks :)
To find the mass of oxygen evolved in the experiment, we need to calculate the difference between the initial mass of potassium chlorate and the mass of potassium chloride residue.
Given:
Mass of potassium chlorate = 1.24 grams
Mass of potassium chloride residue = 0.72 grams
1. Calculate the mass of oxygen evolved:
Mass of oxygen = Mass of potassium chlorate - Mass of potassium chloride residue
= 1.24 grams - 0.72 grams
= 0.52 grams
Therefore, the mass of oxygen evolved in the experiment is 0.52 grams.
To determine the balanced equation for the reaction, we need to know the products formed when potassium chlorate is heated with manganese dioxide. Unfortunately, the question does not provide this information, so we cannot determine the balanced equation.
Moving on to the next part of the question, we need to calculate the percent, by mass, of oxygen based on the given data.
2. Calculate the percent, by mass, of oxygen:
Percent oxygen = (Mass of oxygen / Mass of potassium chlorate) x 100
= (0.52 grams / 1.24 grams) x 100
= 41.94%
Therefore, the percent, by mass, of oxygen based on the given data is approximately 41.94%.