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##
##2KClO_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 gcolor(white)(a/a)|)))##