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?

Question

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

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

HIX Tutor · Accepted Answer

undefined To find the total moles of gases produced, follow these steps:
1. Start with 8 moles of "NaNO"_3.
2. Calculate the moles of gases produced using the percentage yield (30%):
   - Moles of gases = 8 moles of "NaNO"_3 x 30%
                   = 8 moles x 0.30
                   = 2.4 moles of gases will be produced.

HIX Tutor · Answer

undefined First, calculate the moles of product produced using the percentage yield.

1. Moles of product = 8 moles of "NaNO"_3 * 30% = 8 moles * 0.30 = 2.4 moles

2. Since 2 moles of gases are produced in the reaction for every 1 mole of "NaNO"_3 consumed, the total moles of gases produced will be: 2.4 moles of product * 2 = 4.8 moles of gases.

Olivia Anton · Answer

#"4 moles"#

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, this is actually how the chemical equation describing this decomposition reaction should look.

#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#

Essentially, by using the percent yield as a conversion factor, you can determine the reaction's actual yield.

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

As a result, the reaction will produce a total of moles of gases, which will be

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

When the answer is rounded to one significant number, it 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#

Recall that there are two applications for the yield percentage.

Ellie Andrews · Answer

undefined To find the total moles of gases produced, you first need to calculate the theoretical yield of the reaction using the given percentage yield. Then, you can determine the total moles of gases produced based on the stoichiometry of the reaction.

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.

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

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