Question #80f53

The first thing that you need to do here is to find the number of moles of potassium chlorate that undergoes decomposition to produce potassium chloride and oxygen gas.

To do that, use the molar mass of potassium chlorate

#10.3 color(red)(cancel(color(black)("g"))) * "1 mole KClO"_3/(122.55color(red)(cancel(color(black)("g")))) = "0.08405 moles KClO"_3#

Now, you know by looking at the balanced chemical equation that describes this decomposition reaction

#2"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + 3"O"_ (2(g))#

that for every #2# moles of potassium chlorate that undergo decomposition, the reaction produces #3# moles of oxygen gas.

In your case, the reaction will produce

#0.08405 color(red)(cancel(color(black)("moles KClO"_3))) * "3 moles O"_2/(2color(red)(cancel(color(black)("moles KClO"_3)))) = "0.1261 moles O"_2#

Finally, to convert the number of moles to molecules, use Avogadro's constant, which tells you that in order to have #1# mole of oxygen gas, you need to have #6.022 * 10^(23)# molecules of oxygen gas.

#0.1261 color(red)(cancel(color(black)("moles O"_2))) * (6.022 * 10^(23)color(white)(.)"molecules O"_2)/(1color(red)(cancel(color(black)("mole O"_2))))#

# = color(darkgreen)(ul(color(black)(7.59 * 10^(22)color(white)(.)"molecules O"_2)))#

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