Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles?

#"C"_3"H"_8 + "O"_2 - "CO"_2 + "H"_2"O"# (unbalanced)

Question

Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles?
 
#"C"_3"H"_8 + "O"_2 -> "CO"_2 + "H"_2"O"# (unbalanced)

Wyatt Atkins · Accepted Answer

#"0.25 moles C"_3"H"_8#

First, write the chemical equation that describes this combustion reaction in balance.

#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#

Now, the balanced chemical equation tells you that in order for the reaction to produce #3# moles of carbon dioxide, it must consume #1# mole of propane.

This is the theoretical yield of the reaction, i.e. what you would get at #100%# yield.

In your case, the reaction is known to have a percent yield equal to #75%#. This essentially means that for every #100# moles of carbon dioxide that the reaction could theoretically produce, you only get #75# moles.

To convert the sample's mass to moles, use the carbon dioxide molar mass.

#25 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.568 moles CO"_2#

Now, at #100%# yield, the reaction must consume

#0.568 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = "0.189 moles C"_3"H"_8#

in order to produce #0.568# moles of carbon dioxide. However, you know that the reaction has a percent yield equal to #75%#, which means that the reaction produced only #75%# of the number of moles of carbon dioxide that it would have produced at theoretical yield.

This means that in order to get #0.568# moles of carbon dioxide, the reaction must consume enough moles of propane to theoretically produce

#0.568 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("100 moles CO"_2/(75 color(red)(cancel(color(black)("moles CO"_2)))))^(color(blue)("= 75% yield")) = "0.757 moles CO"_2#

Therefore, you can say that the required number of moles of propane is equal to

#0.757 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = color(darkgreen)(ul(color(black)("0.25 moles C"_3"H"_8)))#

The number of sig figs you have for the mass of carbon dioxide is the answer, which is rounded to two sig figs.

Thus, you could say that, in response to this, you have

#"0.189 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.568 moles CO"_2#

#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.757 moles CO"_2#

#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 75% yield")aaa))(->) " " "0.568 moles CO"_2#

Wyatt Adkins · Answer

undefined To calculate the necessary amount of propane (C3H8) in moles, we first need to balance the equation:

C3H8 + 5O2 -> 3CO2 + 4H2O

Now, according to the balanced equation, 1 mole of propane (C3H8) produces 3 moles of carbon dioxide (CO2).

Given that the reaction yield is 75%, we can calculate the actual amount of carbon dioxide produced:

Actual yield = 75% of theoretical yield
Theoretical yield = amount of carbon dioxide desired / yield percentage

Theoretical yield = 25 g / 0.75 = 33.33 g

Now, we can use the molar mass of carbon dioxide to convert grams to moles:

Molar mass of CO2 = 12.01 g/mol (C) + 2(16.00 g/mol) (O) = 44.01 g/mol

Number of moles of CO2 = mass / molar mass = 33.33 g / 44.01 g/mol ≈ 0.757 moles

Since 1 mole of propane produces 3 moles of carbon dioxide, the necessary amount of propane in moles is:

0.757 moles CO2 × (1 mole C3H8 / 3 moles CO2) ≈ 0.252 moles C3H8

Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles? #"C"_3"H"_8 + "O"_2 -&gt; "CO"_2 + "H"_2"O"# (unbalanced)

Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles? #"C"_3"H"_8 + "O"_2 -> "CO"_2 + "H"_2"O"# (unbalanced)