Question #14ef3

1 Answer
Ernest Z.
Apr 16, 2015

The reaction forms 0.0763 mol of #"O"_2#.

Explanation:

The equation is #"2KClO"_3 → "2KCl" + "3O"_2#

Step 1. Calculate the moles of #"KClO"_3#.

#6.23 cancel("g KClO₃") × ("1 mol KClO"_3)/(122.55 cancel("g KClO₃")) = "0.050 84 mol KClO"_3#

Step 2. Use the molar ratio from the equation to calculate the moles of #"O"_2#.

#0.050 84 cancel("mol KClO₃") × ("3 mol O"_2)/(2 cancel("mol KClO₃")) = "0.0763 mol O"_2#

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