How many atoms of oxygen would be produced if you reacted 68.1 g of potassium chlorate according to the following equation: 2 KClO3 → 2KCl + 3O2
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Answer:
The number of moles of KClO
3
is 6.125 g
So,
122.5
6.125
= 0.05 moles.
Two molecules of KClO
3
requires 3 molecules of O
2
Number of moles of O
2
= 3×0.05/2
= 0.075 moles
Volume of 1 mole =22.4L
Volume of O
2
= 22.480.075
= 1.68 L
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