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Question

1.225 g sample of KClO3 was heated under such conditions that a part of it decomposed according to the equations given below. If the amount of O2 evolved was 168 mL at STP, calculate the weight of KClO4.
(Molar mass of KClO3=122.5 g mol1,KClO4=138.5g mol1)KClO3KCl+O2KClO3KClO4+KCl

A
0.875 g
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B
0.519 g
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C
0.250 g
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D
0.315 g
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Solution

The correct option is B 0.519 g
KClO3KCl+O2
Applying the principle of atom conservation (POAC) to O,
Moles of oxygen in KClO3 = Moles of oxygen in O2
3×moles ofKClO3=2moles of O2
3×Weight of KClO3molar mass=2×Volume of O2 (mL) at STP22400
3×WeightofKClO3122.5=2×16822400
Weight of KClO3 = 0.6125 g
The amount of KClO3 left = 1.225 – 0.6125 = 0.6125 g
KClO3KClO4+KCl
Applying the principle of atom conservation (POAC) to O,
Moles of oxygen in KClO3 = Moles of oxygen in KClO4
3 × moles of KClO3 = 4 × moles of KClO4
3×Weight of KClO3molar mass=4×Weight of KClO4molar mass3×0.6125122.5=4×Weight of KClO4138.5Weight of KClO4=0.519 g

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