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Grade: 9
        
A 1g sample of KClO3 was heated under such conditions that a part of it decomposed according to the following equation:
2KClO3 -> 2KCl+3O2
The remaining part underwent change according to the following equation:
4KClO3 -> 3KClO4+KCl
If amount of O2 evolved was 146.8ml at NTP, calculate the percent by mass of KClO4 in the residue.
5 months ago

Answers : (2)

Arun
24716 Points
							
2 KClO3 ? 2 KCl + 3 O2

(0.1458 L O2) / (22.414 L/mol) x (2/3) x (122.5498 g KClO3/mol) = 
0.5314 g KClO3 participated in the first reaction 

1 g - 0.5314 g = 0.4686 g KClO3 decomposed according to the second reaction

4 KClO? 3 KClO4 + KCl

(0.4686 g KClO3) / (122.5498 g KClO3/mol) x (3/4) x (138.5492 g KClO4/mol) = 0.3973 g KClO4 formed in the second reaction
5 months ago
Vikas TU
11682 Points
							
Dear student 
The above ans is incorrect
Moles of   KClO3 = 1/122.55 g/mol = 0.081 mol
(1) 2KClO3 - 2KCl + 3O2 
Moles of oxygen gas at 1 atm and 273 K can be calculated by using ideal gas equation :
n = PV/RT 
n = 0.00499 mol = Moles of oxygen gas produced at given conditions
According to reaction(1), 2 moles of  produces 3 moles of oxygen.
Then 0.00499 mol of oxygen will be produced from:
 of 
2/3 * 0.0499 mol = 0.00332 mol of KClO3 
Moles of  used in part(1) = 0.00332 mol
(2) 4KClO3 - 3KclO4 +Kcl 
Moles of  used in part(2)=
= 0.0081 mol - 0.00332 mol = 0.00478 mol
According to reaction (2), 4 moles of KClO3  gives 3 moles of Kcl O4  .
Then 0.00478 moles of KClO3  will give:
 3/4 * 0.00478 mol = 0.003585 mol of KCl O3
Mass of 0.003585 moles of:
0.003585 mole × 138.55 g/mol =0.4967 g
Percentage by weight of  in the residue:
% = Mass of KClO4 / Mass of KClO3
= 49%
 
The % by weight of  in the residue 49.67 %
 
5 months ago
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