what is the volume of oxygen at STP liberated by heating 12.25 gram of potassium chlorate
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The equation is
2KClO3---->2KCl + 3O2
2 × 122.5 gives 3 × 22.4 liters of O2.
12.25 gives y liters
y = 12.25 × 3 × 22.4/ 2 × 122.5
y = 0.1 × 3 × 11.2
y = 3.36 liters of O2.
2KClO3---->2KCl + 3O2
2 × 122.5 gives 3 × 22.4 liters of O2.
12.25 gives y liters
y = 12.25 × 3 × 22.4/ 2 × 122.5
y = 0.1 × 3 × 11.2
y = 3.36 liters of O2.
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