# 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)

Sep 24, 2015

${\text{165 g KClO}}_{3}$

#### Explanation:

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

$\textcolor{red}{2} {\text{KClO"_text(3(s]) -> 2"KCl"_text((s]) + color(blue)(3)"O}}_{\textrm{\left(g\right]}}$

Notice that you have a $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio between potassium chlorate, ${\text{KClO}}_{3}$, and oxygen gas, ${\text{O}}_{2}$.

This means that for a reaction that has an 100% percent yield, 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 $\text{42.0 g}$ 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% percent yield?

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.

$\text{% yield" = "actual yield"/"theoretical yield} \times 100$

You know that the reaction's actual yield is 42.0 g of oxygen gas, which means that the theoretical yield must be

$\text{65.0%" = "42.0 g"/"theoretical yield} \times 100$

$\text{theoretical yield" = ("42.0 g" * 100)/65 = "64.6 g}$

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 the mole ratio that exists between the two 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 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)

Sep 24, 2015

You need 165 g of ${\text{KClO}}_{3}$.

#### Explanation:

$2 {\text{KClO"_3 → "2KCl" + "3O}}_{2}$

From the above reaction it can be seen that

${\text{2 mol of KClO"_3 = "3 mol of O}}_{2}$

${\text{1 mol of KClO"_3 = 3/2 " mol O"_2 = "1.5 mol O}}_{2}$.

The yield of the reaction is given in %, i.e., 65 %.

${\text{1 mol KClO}}_{3}$ generates ${\text{1.5 mol O}}_{2}$

Since the yield is 65 %, the number of moles of ${\text{O}}_{2}$ generated per mole of ${\text{KClO}}_{3}$ is

$\text{1.5 mol" xx 65/100 = "0.975 mol}$

$\text{Number of moles of O"_2 = "42.0 g"/"32.0 g/mol" = "1.31 mol}$

${\text{Number of moles of KClO"_3 = "1.31 mol O"_2//("0.975 mol O"_2//"mol KClO"_3) = "1.35 mol KClO}}_{3}$

${\text{Weight of KClO"_3 = "1.35 mol KClO"_3xx"Molecular weight of KClO}}_{3}$

$= \text{1.34 mol"xx"122.6 g/mol = 165 g}$