# When heated, KClO3 decomposes into KCl and O2. If this reaction produced 20.3 g of KCl, how much O2 was produced (in grams)?

##”13.1 g O”_2##
Start by writing the balanced chemical equation that describes the 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)## that exists between the two products of the reaction to a gram ratio by using the molar masses of the two .
##”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 , 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)|)))##

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