Question #186eb

Start by writing the balanced chemical equation that describes the decomposition of potassium chlorate, #"KClO"_3#, to form potassium chloride, #"KCl"#, and oxygen gas, #"O"_2#

#2"KClO"_text(3(s]) -> color(red)(2)"KCl"_text((s]) + color(blue)(3)"O"_text(2(g]) uarr#

Notice that for every two moles of potassium chlorate that undergo decomposition, you get #color(red)(2)# moles of potassium chloride and #color(blue)(3)# moles of oxygen gas.

You can convert this #color(red)(2):color(blue)(3)# mole ratio that exists between the two products of the reaction to a gram ratio by using the molar masses of the two compounds.

#"For KCl": " " M_M = "74.55 g mol"^(-1)#

#"For O"_2: " " M_M = "32.0 g mol"^(-1)#

This tells you that every mole of potassium chloride has a mass of #"74.55 g"#, which implies that #color(red)(2)# moles will have a mass of

#2 color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "149.1 g"#

For oxygen gas, every mole has a mass of #"32.0 g"#, which means that #color(blue)(3)# moles will have a mass of

#3 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "96.0 g"#

The #color(red)(2):color(blue)(3)# mole ratio will be equivalent to a #149.1 : 96.0# gram ratio, i.e. you get #"96.0 g"# of oxygen gas for every #"149.1 g"# of potassium chloride produced by the reaction.

This means that #"20.3 g"# of potassium chloride would correspond to

#20.3color(red)(cancel(color(black)("g KCl"))) * "96.0 g O"_2/(149.1color(red)(cancel(color(black)("g KCl")))) = "13.07 g O"_2#

Rounded to three sig figs, the number of sig figs you have for the mass of potassium chloride, the answer will be

#m_(O_2) = color(green)(|bar(ul(color(white)(a/a)"13.1 g"color(white)(a/a)|)))#