Question

If 195 grams of Potassium Chlorate (KClO3) decomposes, how much oxygen (O2) will be produced in this reaction?

Answer 1

First, we have to convert from 195 grams of KClO3 to moles, using its molar mass which you can calculate it using the periodic table (122.5 g/mol):

`195gKClO_3\cdot\frac{1\text{ mol }KClO_3}{122.5\text{ g }KClO_3}=1.59\text{ mol }KClO_3.`

Now that we have this value, we can find the number of moles produced of oxygen doing a rule of three taking into account that 2 moles of KClO3 produces 3 moles of oxygen:

`\begin{gathered} 2\text{ moles }KClO_3\to3molesO_2 \\ 1.59\text{ moles }KClO_3\to?molesO_2 \end{gathered}`

And the calculation would be:

`1.59\text{ moles }KClO_3\cdot\frac{3molesO_2}{2\text{ moles }KClO_3}=2.39molesO_2.`

The final step is to convert 2.39 moles of oxygen to grams using its molar mass. The molar mass of oxygen (O2) is 32 g/mol:

`2.39molesO_2\cdot(32gO_2)/(1molO_2)=76.4gO_2.`

The answer is that 195 grams of KClO3 decomposed, produces 76.4 grams of oxygen (O2).