##165 g KClO_3##
Once again start with the balanced chemical equation for this
##color(red)(2)KClO_text(3(s]) -> 2KCl_text((s]) + color(blue)(3)O_text((g])##
Notice that you have a ##color(red)(2):color(blue)(3)## between potassium chlorate ##KClO_3## and oxygen gas ##O_2##.
This means that for a reaction that has an 100% every two moles of potassium chlorate will produce three moles of oxygen gas.
Keep this in mind.
So you know that your reaction must produce ##42.0 g## of oxygen gas. Use oxygen gas’ molar mass to find how many moles must be produced
##42.0color(red)(cancel(color(black)(g))) * (1 mole O_2)/(32.0color(red)(cancel(color(black)(g)))) = 1.3125 moles O_2##
So how many moles of potassium chlorate would you need If the reaction had an 100% ?
Use the aforementioned mole ratio to find
##1.3125color(red)(cancel(color(black)(moles O_2))) * (color(red)(2) moles KClO_3)/(color(blue)(3)color(red)(cancel(color(black)( moles O_2)))) = 0.875 moles KClO_3##
However you know for a fact that the percent yield of the reaction is not 100% but 65.0%. This means that you will need to use more potassium chlorate to produce this much oxygen gas.
Percent yield is defined as the actual yield of the reaction divided by the theoretical yield of the reaction.
##% yield = actual yield/theoretical yield xx 100##
You know that the reaction’s actual yield is 42.0 g of oxygen gas which means that the theoretical yield must be
##65.0% = 42.0 g/theoretical yield xx 100##
##theoretical yield = (42.0 g * 100)/65 = 64.6 g##
This means that you need to find how many grams of potassium chlorate would theoretically produce 64.6 g of oxygen gas.
Once again use oxygen’s molar mass and ratio that exists between the two
##64.6color(red)(cancel(color(black)(g O_2))) * (1 mole O_2)/(32.0color(red)(cancel(color(black)(g O_2)))) = 2.019 moles O_2##
This means that you need to use
##2.019color(red)(cancel(color(black)(moles O_2))) * (color(red)(2) moles KClO_3)/(color(blue)(3)color(red)(cancel(color(black)( moles O_2)))) = 1.346 moles KClO_3##
moles of potassium chlorate. Finally use the compound’s molar mass to find how many grams would contain this many moles
##1.346color(red)(cancel(color(black)(moles KClO_3))) * 122.55 g/(1color(red)(cancel(color(black)(mole KClO_3)))) = color(green)(165 g KClO_3)##