Calculate the mass of potassium chlorate required to liberate 6.72dmcube of oxygen at step.Molar mass of potassium chlorate is 122.5g mol inverse
Ans.24.5g
Asked by aniketbele4084
| 10th Aug, 2019,
12:00: AM
Expert Answer:
Given:
Volume of oxygen liberated = 6.72 dm3
1 mole of oxygen = 22.4 dm3 = 1mole
6.72 dm3 of oxygen will be
The reaction can be written as,
To liberate 3 mole of oxygen 2 mole of KClO3 are required
So 1 mole of oxygen will require 
So for 0.3 mole of oxygen will require 0.66×0.3 = 0.2 mol of KClO3
We have,
Molar mass of KClO3 = 122.5 g/mol
1 mole of KClO3 = 122.5 g
0.2 mole of KClO3 = 0.2×122.5
= 24.5 g
The mass of KClO3 required to liberate 6.72 dm3 of oxygen is 24.5 g
Answered by Varsha
| 12th Aug, 2019,
11:45: AM
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