What volume of oxygen, o2, at stp can be prepared by the complete decomposition of 0.100 mol of potassium chlorate kclo3? 2kclo3=2kcl+3o2

Question

I need help on this question. I would like work to help me figure out the problem.

Answer 1

The very first thing you should do in problems of this type is to write the balanced equation (which in this case is already given to you):

2KClO₃ ==> 2KCl + 3O₂(g)

Next, examine the stoichiometry which shows you that it takes TWO moles of KClO₃ to produce THREE moles of O₂(g).

Now, use this mole ratio to calculate moles of O₂ formed from 0.100 moles of KClO₃

0.100 moles KClO₃ x 3 moles O₂/2 moles KClO₃ = 0.150 moles O₂ formed

 

Finally, convert moles of O₂ to volume at STP:

At STP each mole of an ideal gas occupies 22.4 liters.  Thus, 0.150 moles O₂ x 22.4 liters/mole = 3.36 liters O₂