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3.0 mol sample of KClO3 was decomposed according to the equation 2KClO3 = 2KCl + 3 O2 How many moles of O2 are formed assuming 100% yield? is it 3 moles?

I don't want to work through the details of this problem completely. Tell me what you don't understand about it.

I don't understand how to set up the equation. I do know that 6022 times 10 to the 23 power is a mole but I don't know how to get the answer when we are using 3 moles

3.0 mol sample of KClO3 was decomposed according to the equation 2KClO3 = 2KCl + 3 O2 How many moles of O2 are formed assuming 100% yield? is it 3 moles?

I think you are making the problem harder than it is.
Write the equation.
2KClO3 ==> 2KCl + 3O2.

So the equation tells us 2 mols KClO3 will produce 3 mols O2. So how much will 1 mol KClO3 produce. Obviously that is mols O2 = 3 mols O2 x (1 mol KClO3/2 mols KClO3) = 1.5 mols O2 OR we produce 1.5 mols O2 for every mol KClO3 decomposed. So decomposing 3 mols will just be 3 times that or 3 x 1.5 = 1.5 mols. To put all of that in equation form,
mols O2 = 3 mols O2 x (3 mols KClO3/2 mols KClO3) = 9/2 = 4.5 mols O2.

If you wish to set up a proportion, just do the following.
2mol KClO3/3 mols O2 = 3 mols KClO3/x mols O2. Solve for x.

2/3 = 3/x
2x = 9
x = 4.5

  • chem -

    2.85 mol O2

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