Question #dd20a

The first thing to do here is to write a balanced chemical equation that describes this decomposition reaction

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

Notice that it takes #color(red)(2)# moles of potassium chlorate to produce #color(blue)(3)# moles of oxygen gas.

This means that if you know how many moles of oxygen gas were produced by the reaction, you can backtrack and use this #color(red)(2):color(blue)(3)# mole ratio to figure out how many moles of potassium chlorate underwent decomposition.

Now, STP conditions, which were probably given to you as a pressure of #"1 atm"# and a temperature of #0^@"C"#, are characterized by the fact that one mole of any ideal gas occupies #"22.4 dm"^3# -- this is known as the molar volume of a gas at STP.

Use this value to calculate how many moles of oxygen gas were produced by the reaction

#6.72 color(red)(cancel(color(black)("dm"^3))) * " mole O"_2/(22.4color(red)(cancel(color(black)("dm"^3)))) = "0.30 moles O"_2#

This means that the decomposition reaction consumed

#0.30 color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)color(white)(a)"moles KClO"_3)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = "0.20 moles KClO"_3#

Now all you have to do is use the molar mass of potassium chlorate to calculate how many grams would contain this many moles

#0.20 color(red)(cancel(color(black)("moles KClO"_3))) * "122.5 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("24.5 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.