Answer:
64.2 L
Step-by-step explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature, and Pressure.
Under these conditions for pressure and temperature, one mole of any ideal gas occupies 22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this:
2KClO3(s] −−−→2KCl(s]+3O2(g]↑
Notice that you have a 2/3 mole ratio between potassium chlorate and oxygen gas. This tells you that the reaction will always produce 3/2 times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that 231-g sample.
231g ⋅ 1 mole KClO3/ 122.55g = 1.885 moles KClO3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885 moles KClO3 ⋅ 3 moles O2/2 moles KClO3 = 2.8275 moles O2
So, what volume would this many moles occupy at STP?
2.8275 moles ⋅ 22.7 L/ 1mol = 64.2 L
This was alotta info but i hope it helped