Question #c0e53

Your starting point here will be the balanced chemical equation for this decompsition reaction

#2"KClO"_text(3(s]) stackrel(color(red)("heat")color(white)(xx))(->) 2"KCl"_text((s]) + 3"O"_text(2(g]) uarr#

Notice that you have a #2:3# mole ratio between potassium chlorate and oxygen gas. This means that the rection will always produce #3/2# times more moles of oxygen than you had moles of potassium chlorate that took part in the reaction.

Now, let's assume that you have a sample of potassium chlorate of #x# grams.

The number of moles of potassium chlorate can be determined using its molar mass

#xcolor(red)(cancel(color(black)("g"))) * "1 mole KClO"_3/(122.5color(red)(cancel(color(black)("g")))) = x/122.5"moles KClO"_3#

This means that the reaction produced

#x/122.5color(red)(cancel(color(black)("moles KClO"_3))) * "3 moles O"_2/(2color(red)(cancel(color(black)("moles KClO"_3)))) = (3x)/245"moles O"_2#

To get the mass of oxygen produced by the reaction, use oxygen gas' molar mass

#(3x)/245color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = (96x)/245"g O"_2#

This means that the percentage loss in the mass of the sample will be

#"% loss" = m_"loss"/m_"initial" xx 100#

#"% loss" = (96/245xcolor(red)(cancel(color(black)("g"))))/(xcolor(red)(cancel(color(black)("g")))) xx 100 = 96/245 xx 100 = color(green)(39.2%)#

The sample will be #39.2%# lighter after the decomposition reaction.