calculate the mass of Potassium chlorate required to liberate the 6. 72 decimeter cube of oxygen at STP molar mass of the potassium chloride is 122. 5 gram per mole
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hloo frnd and your answer is as follows........
The reaction is as follows:
2KClO3 → 2KCl + 3O2
Thus, we can see that
2mol KClO3 will produce 3 mol O2
At STP 1 mol O2 will occupy a volume of 22.4L
Given, 6.72 L of O2 is liberated
6.72L = 6.72 / 22.4
= 0.3 mol O2
This will surely require 0.2 mol KClO3
Hence,
Mass KClO3 required = 0.2 x 122.5 = 24.5g
hope it may helps u....☺☺
The reaction is as follows:
2KClO3 → 2KCl + 3O2
Thus, we can see that
2mol KClO3 will produce 3 mol O2
At STP 1 mol O2 will occupy a volume of 22.4L
Given, 6.72 L of O2 is liberated
6.72L = 6.72 / 22.4
= 0.3 mol O2
This will surely require 0.2 mol KClO3
Hence,
Mass KClO3 required = 0.2 x 122.5 = 24.5g
hope it may helps u....☺☺
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