Question #badeb

Start by writing the balanced chemical equation that describes this decomposition reaction.

#2"KNO"_ (3(s)) -> 2"KCl"_ ((s)) + 3"O"_ (2(g)) uarr#

As you can see, the reaction produces #3# moles of oxygen gas whenever #2# moles of potassium chlorate undergo decomposition.

This tells you that if you divide the number of moles of oxygen gas produced by the reaction and the number of moles of potassium chlorate that underwent decomposition, you will end up with #3/2#.

Use the molar mass of potassium chlorate to convert the number of grams to moles

#49.9 color(red)(cancel(color(black)("g"))) * "1 mole KClO"_3/(122.55color(red)(cancel(color(black)("g")))) = "0.4072 moles KClO"_3#

You can now use the mole ratio that exists between oxygen gas and potassium chlorate to say that you will have

#"moles of O"_2/("0.4072 moles KNO"_3) = "3 moles O"_2/"2 moles KClO"_3#

This will get you

#"moles of O"_2 = (0.4072 color(red)(cancel(color(black)("moles KClO"_3))))/(2color(red)(cancel(color(black)("moles KClO"_3)))) * "3 moles O"_2#

#"moles of O"_2 = color(darkgreen)(ul(color(black)("0.612 moles O"_2)))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of potassium chlorate.