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Chemistry 12 Online
OpenStudy (anonymous):

If this reaction produced 75.4g of KCl, how much of O2 was produced (in grams)

OpenStudy (anonymous):

reaction please???

OpenStudy (anonymous):

2KClO3=2KCl+3O2

OpenStudy (anonymous):

n(KCl)=m/M =75.4/(39.1+35.45) =1.011mol n(O2)=3/2n(KCl) =1.517mol m(O2)= nM = 1.517 x (16.00x2) =48.5g

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