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
 
equals space fraction numerator 6.72 over denominator 22.4 end fraction

equals 0.3 space mol
 
The reaction can be written as,
 
2 KClO subscript 3 subscript open parentheses straight s close parentheses end subscript end subscript space rightwards arrow with increment space space space space space on top space space 2 KCl subscript open parentheses straight s close parentheses end subscript space space space plus space space space 3 straight O subscript 2 subscript open parentheses straight g close parentheses end subscript end subscript
2 space mol space space space space space space space space space space space space space space space space 2 space mol space space space space space space space space space space space 3 space mol
 
To liberate 3 mole of oxygen 2 mole of KClO3 are required
 
So 1 mole of oxygen will require 2 over 3 space equals 0.66 space mol space of space KClO subscript 3
 
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