If the percent yield for the following reaction is 65.0%, how many grams of KClO3 are needed to produce 42.0 g of O2? 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

##”165 g KClO”””_3##
Once again, start with the balanced chemical equation for this
##color(red)(2)”KClO”_text(3(s]) -> 2″KCl”_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)##