A common laboratory preparation for small for quantities of #O_2# is to decompose #KClO_3# by heating: #2KClO_3 (s) - 2KCl (s) + 3O_2 (g)#. If the decomposition of 2.00 g of #KCIO_3# gives 0.720 g of #O_2#, what is the percent yield for the reaction?

Question

A common laboratory preparation for small for quantities of #O_2# is to decompose #KClO_3# by heating: #2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)#. If the decomposition of 2.00 g of #KCIO_3# gives 0.720 g of #O_2#, what is the percent yield for the reaction?

Cameron Andrews · Accepted Answer

#91.9%#

To begin, look at the balanced chemical equation that explains this breakdown process.

#color(darkgreen)(2)"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(blue)(3)"O"_ (2(g))#

Notice that it takes #color(darkgreen)(2)# moles of potassium chlorate to produce #color(blue)(3)# moles of oxygen gas. This represents the reaction's theoretical yield, i.e. the amount of oxygen gas that you can expect to see for a reaction that has a #100%# yield.

Use the molar masses of potassium chlorate and oxygen gas to convert the mole ratio to a gram ratio because the reaction gives you grams of potassium chlorate.

#(color(darkgreen)(2)color(red)(cancel(color(black)("moles KClO"_3))) * "122.55 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))))/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = "2.553 g KClO"_3/"1 g O"_2#

This means that for every #"2.553 g"# of potassium chlorate that undergo combustion, the reaction can theoretically produce #"1 g"# of oxygen gas.

Now, the problem tells you that you decompose #"2.00 g"# of potassium chlorate. The reaction should theoretically produce

#2.00 color(red)(cancel(color(black)("g KClO"_3))) * "1 g O"_2/(2.553color(red)(cancel(color(black)("g KClO"_3)))) = "0.7834 g O"_2#

This represents the reaction's theoretical yield. You know that the reaction actually produced #"0.720 g"# of oxygen gas, which means that its percent yield was

#(0.720 color(red)(cancel(color(black)("g O"_2))))/(0.7834color(red)(cancel(color(black)("g O"_2))))xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(91.9%)color(white)(a/a)|)))#

Three sig figs are used to round the result.

Noah Aldrich · Answer

undefined To find the percent yield for the reaction, we first need to calculate the theoretical yield of oxygen (O₂) based on the given mass of potassium chlorate (KClO₃), and then compare it to the actual yield.

1. Calculate the molar mass of KClO₃:
   K: 39.10 g/mol
   Cl: 35.45 g/mol
   O: 16.00 g/mol
   Total molar mass = 39.10 + 35.45 + (3 * 16.00) = 122.55 g/mol

2. Convert the given mass of KClO₃ to moles:
   Moles of KClO₃ = mass / molar mass
                  = 2.00 g / 122.55 g/mol
                  ≈ 0.016 mol

3. According to the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂. So, calculate the theoretical yield of O₂:
   Theoretical moles of O₂ = (moles of KClO₃ * 3) / 2
                           = (0.016 mol * 3) / 2
                           = 0.024 mol

4. Calculate the theoretical mass of O₂:
   Theoretical mass of O₂ = moles * molar mass
                           = 0.024 mol * (2 * 16.00 g/mol)
                           = 0.768 g

5. Calculate the percent yield:
   Percent yield = (actual yield / theoretical yield) * 100%
                 = (0.720 g / 0.768 g) * 100%
                 ≈ 93.75%

So, the percent yield for the reaction is approximately 93.75%.

A common laboratory preparation for small for quantities of #O_2# is to decompose #KClO_3# by heating: #2KClO_3 (s) -&gt; 2KCl (s) + 3O_2 (g)#. If the decomposition of 2.00 g of #KCIO_3# gives 0.720 g of #O_2#, what is the percent yield for the reaction?

A common laboratory preparation for small for quantities of #O_2# is to decompose #KClO_3# by heating: #2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)#. If the decomposition of 2.00 g of #KCIO_3# gives 0.720 g of #O_2#, what is the percent yield for the reaction?