Oxygen can be prepared by the decomposition of potassium chlorate(KClO3)
Q: 200 g of O2 were obtained. How many moles of KClO3 would have decomposed?
Answers
Answered by
1
Answer:
I think 1 molecule of kclo3 was decomposed
Answered by
1
Answer:
the answer is 39.2 mole
Explanation:
kclo3=39+35.5+3×16=122.5
O2=2×16=32
32 grm O2 = 1 mol
200 grm O2= 1/32×200
=3.125
122.5 grm kclo3 = 1 mol
3.125 grm kclo3 = 1/122.5×3.125
=39.2mole
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