If the percentage yield of the given reaction is #30%#, how many total moles of the gases will be produced if #8# moles of #"NaNO"_3# are initially taken?

#4"NaNO"_3 -> 2"NaNO"_2+2"N"_2+5"O"_2#

NOTE: This is the chemical equation given by the student.

1 Answer
Jul 10, 2018

Answer:

#"4 moles"#

Explanation:

The idea here is that a reaction's percent yield tells you how many moles of a product will actually be produced by a chemical reaction for every #100# moles of this product that could theoretically be produced by the reaction.

In your case, the reaction is said to have a percent yield of #30%#. This means that for every #100# moles of a product that the reaction could produce, you will only get #30# moles.

Now, the chemical equation that describes this decomposition reaction should actually look like this

#4"NaNO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"Na"_ 2"O"_ ((s)) + 2"N"_ (2(g)# #uarr + 5"O"_ (2(g)) uarr#

You know that for this reaction, you use #8# moles of sodium nitrate. According to the mole ratios that exist between the reactant and the products, this reaction should theoretically produce

#8 color(red)(cancel(color(black)("moles NaNO"_3))) * ("2 moles Na"_2"O")/(4color(red)(cancel(color(black)("moles NaNO"_3)))) = "4 moles Na"_2"O"#

#8 color(red)(cancel(color(black)("moles NaNO"_3))) * "2 moles N"_2/(4color(red)(cancel(color(black)("moles NaNO"_3)))) = "4 moles N"_2#

#8color(red)(cancel(color(black)("moles NaNO"_3))) * "5 moles O"_2/(4color(red)(cancel(color(black)("moles NaNO"_3)))) = "10 moles O"_2#

These values represent the theoretical yield of the reaction, i.e. what you would expect to see for a #100%# yield.

In your case, the #30%# yield means that the actual yield of the reaction will be

#4 color(red)(cancel(color(black)("moles Na"_2"O"))) * ("30 moles Na"_2"O")/(100color(red)(cancel(color(black)("moles NaNO"_3)))) = "1.2 moles Na"_2"O"#

#4 color(red)(cancel(color(black)("moles N"_2))) * "30 moles N"_2/(100color(red)(cancel(color(black)("moles N"_2)))) = "1.2 moles N"_2#

#10color(red)(cancel(color(black)("moles O"_2))) * "30 moles O"_2/(100color(red)(cancel(color(black)("moles O"_2)))) = "3 moles O"_2#

Basically, you can find the actual yield of the reaction by using the percent yield as a conversion factor.

#"theoretical yield" * overbrace("actual yield"/"100 moles as a theoretical yield")^(color(blue)("the percent yield")) = "actual yield"#

Therefore, the total number of moles of gases that will be produced by the reaction will be

#"1.2 moles N"_2 + "3 moles O"_2 = "4.2 moles gases"#

Rounded to one significant figure, the answer will be

#"moles of gases" = color(darkgreen)(ul(color(black)("4 moles")))#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaa)#

Notice that you can get the same result by assuming that out of the #8# moles of sodium nitrate that are available, only #30%# actually take part in the reaction.

#8 color(red)(cancel(color(black)("moles NaNO"_3))) * ("30 moles NaNO"_3 \ "react")/(100color(red)(cancel(color(black)("moles NaNO"_3 \ "available")))) = "2.4 moles NaNO"_3#

This means that your reaction will use up #2.4# moles of sodium nitrate at #100%# yield to produce

#2.4 color(red)(cancel(color(black)("moles NaNO"_3))) * "2 moles naNO"_2/(4color(red)(cancel(color(black)("moles NaNO"_3)))) = "1.2 moles NaNO"_2#

#2.4 color(red)(cancel(color(black)("moles NaNO"_3))) * "2 moles N"_2/(4color(red)(cancel(color(black)("moles NaNO"_3)))) = "1.2 moles N"_2#

#2.4color(red)(cancel(color(black)("moles NaNO"_3))) * "5 moles O"_2/(4color(red)(cancel(color(black)("moles NaNO"_3)))) = "3 moles O"_2#

So remember, you can use the percent yield in two ways.

  1. Assume that all the moles of the reactant take part in the reaction and use the percent yield to find the actual yield
  2. Assume that not all the moles of the reactant take part in the reaction, use the percent yield to find the number of moles of the reactant that actually take part in the reaction, and use an #100%# yield to find the actual yield