Once again, start with the balanced chemical equation because that this decomposition reaction

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

Notice that you have actually a #color(red)(2):color(blue)(3)# mole ratio in between potassium chlorate, #"KClO"""_3#, and oxygen gas, #"O"""_2#.

This way that because that a reaction that has actually an 100% percent yield, every 2 moles that potassium chlorate will develop three moles of oxygen gas.

Keep this in mind.

So, you recognize that her reaction must create #"42.0 g"# the oxygen gas. Usage oxygen gas" molar mass to find how plenty of moles must be produced

#42.0color(red)(cancel(color(black)("g"))) * ("1 mole O"""_2)/(32.0color(red)(cancel(color(black)("g")))) = "1.3125 mole O"""_2#

So, how numerous moles of potassium chlorate would certainly you require If the reaction had an 100% percent yield?

Use the abovementioned 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)(" mole O"""_2)))) = "0.875 moles KClO"""_3#

However, you understand for a reality that the percent yield of the reaction is not 100%, yet 65.0%. This means that girlfriend will should use more potassium chlorate to develop this lot oxygen gas.

Percent productivity is identified as the actual yield the the reaction split by the theoretical yield of the reaction.

#"% yield" = "actual yield"/"theoretical yield" xx 100#

You recognize that the reaction"s actual yield is 42.0 g of oxygen gas, which means that the theoretical yield should be

#"65.0%" = "42.0 g"/"theoretical yield" xx 100#

#"theoretical yield" = ("42.0 g" * 100)/65 = "64.6 g"#

This way that you need to discover how many grams of potassium chlorate would theoretically create 64.6 g of oxygen gas.

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Once again, use oxygen"s molar mass and also the mole ratio that exists in between the 2 compounds

#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 way that you have to use

#2.019color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)" mole KClO"""_3)/(color(blue)(3)color(red)(cancel(color(black)(" moles O"""_2)))) = "1.346 mole KClO"""_3#

moles that potassium chlorate. Finally, use the compound"s molar fixed to find how numerous grams would contain this numerous 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)#