How many grams of oxygen can be produced if you have 5.0 grams of KCLO3 in the following reaction? 2KCLO3 -> 2KCL + 3O2

1 mol of potassium chlorate gives 3 moles of oxygen from the equation. This can be written as 3 mol O2 / 1 mol KClO3 (this is a stoichiometric ratio)

Convert g of KClO3 to moles:
Given 5 g of KClO3. This is same as
5 g KClO3 * (1 mol KClO3/122.55 g); g cancels out in the numerator and in the denominator, to yield (5/122.55) mol or 0.040799674 mol.

To obtain moles of O2 released, we will use the stoichiometric ratio we established at the top.

Use stoichiometry to calculate moles of O2:
(3 mol O2/1 mol KClO3)* 0.040799674 mol KClO3 (mol KClO3 cancelsl out in the numerator and denominator) yielding 0.122399021 mol O2.

Use molar mass of O2 to convert mol to g:
32 g O2/1mol O2 (molar mass) from the periodic table. Each O weights 16 g/mol; O2 has two atoms, thus the molar mass would be 32 g/mol.

0.122399021 mol O2 *(32 g/mol O2) (mol O2 cancels out in the numerator and in the denominator, yielding 3.91679.

In the question the mass has two significant figures; thus the mass of O2 released would be 4.0 g (two significant figures).

Method 2

You can see from the equation that
122.55 g of KClO3 yields 3*32 g of O2.
1 g of KClO3 would yield 0.783353733 g of O2.

Thus 5 g of KClO3 would yield 5*0.783353733 g of O2 or 4.0 g O2.