calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP. Molar mass of KC1O3 is 122.5g mol-1
Answer (1)
20th Oct, 2020
Hi
Since 2 moles of KClO3 yields 2 moles of KCl and 3 moles of O2
Therefore, the mole ratio for the reaction is 2:2:1
According to the information provided in the question,
By using ideal gas law, 22.4l = 1 mole
Then, 6.72l= 1×6.72/22.4 = 0.3 moles
As per the mole ratio, 0.3 moles of oxygen to be liberated, we need 0.3×2=0.6 moles of KClO3
Moles= mass/molar mass
Molar mass of potassium chlorate is 122.5g/mol that is given and moles is 0.6
Hence, mass of KClO3 required = 122.5g/mol×0.6 moles = 73.5g
Thank you
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