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)

Question

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)

Wyatt Atkins · Accepted Answer

You need 165 g of #"KClO"_3#.

#2"KClO"_3 → "2KCl" + "3O"_2# It is evident from the response above that #"2 mol of KClO"_3 = "3 mol of O"_2# #"1 mol of KClO"_3 = 3/2 " mol O"_2 = "1.5 mol O"_2#. The reaction's yield, or 65%, is expressed in percentage terms. #"1 mol KClO"_3# generates #"1.5 mol O"_2# Since the yield is 65 %, the number of moles of #"O"_2# generated per mole of #"KClO"_3# is #"1.5 mol" xx 65/100 = "0.975 mol"# #"Number of moles of O"_2 = "42.0 g"/"32.0 g/mol" = "1.31 mol"# #"Number of moles of KClO"_3 = "1.31 mol O"_2//("0.975 mol O"_2//"mol KClO"_3) = "1.35 mol KClO"_3# #"Weight of KClO"_3 = "1.35 mol KClO"_3xx"Molecular weight of KClO"_3# #= "1.34 mol"xx"122.6 g/mol = 165 g"#

Caleb Adams · Answer

#"165 g KClO"""_3#

Start with the balanced chemical equation for this breakdown reaction once more.

#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)# mole ratio between potassium chlorate, #"KClO"""_3#, and oxygen gas, #"O"""_2#.

This indicates that every two moles of potassium chlorate will result in the production of three moles of oxygen gas in a reaction with a 100% yield.

Remember this.

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#

If the reaction yielded 100%, how many moles of potassium chlorate would you need?

Utilize the mole ratio mentioned above to determine

#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#

You are aware, however, that the reaction's percent yield is 65.0%, not 100%, which indicates that more potassium chlorate will be required in order to generate this amount of oxygen gas.

The real yield of the reaction divided by the theoretical yield of the reaction is the definition of percent yield.

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

Since the actual yield of the reaction is 42.0 g of oxygen gas, you can infer 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 implies that you must calculate the number of grams of potassium chlorate that, in theory, would yield 64.6 grams of oxygen gas.

Utilize the molar mass of oxygen and the mole ratio between the two compounds once more.

#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 implies that you must make use of

#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. Lastly, determine how many grams this many moles would contain using the compound's molar mass.

#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)#

Axel Alvarado · Answer

undefined To calculate the grams of KClO3 needed, use the stoichiometry of the reaction. If the percent yield is 65.0%, you would need to use the formula:

$$ \text{Grams of KClO3 needed} = \frac{\text{Grams of O2 desired}}{\text{Percent yield} \times \text{Stoichiometric coefficient of O2}} $$

Substitute the values:

$$ \text{Grams of KClO3 needed} = \frac{42.0 \, \text{g}}{0.65 \times 3} $$