Question #a24ba

1 Answer
Aug 14, 2016

Here's what I got.


The idea here is that sodium bicarbonate, #"NaHCO"_3#, undergoes decomposition at a much lower temperature than sodium carbonate, #"Na"_2"CO"_3#, to produce sodium carbonate, water, and carbon dioxide

#color(darkgreen)(2)"NaHCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "Na"_ 2"CO"_ (3(s)) + "CO"_ (2(g)) + "H"_ 2"O"_ ((g))#

As you can see, heating the mixture, usually at around #150^@"C"# to get a fast decomposition, will produce carbon dioxide and water vapor.

These two products represent the mass lost upon heating. In other words, when you heat the sodium bicarbonate, carbon dioxide and water vapor are given off, leaving behind solid sodium carbonate.

Now, you know that this mass loss amounts to #"2.28 g"#. This means that you have

#m_("CO"_ 2) + m_("H"_ 2"O") = "2.28 g" " " " "color(orange)("(*)")#

Now, let's say that the reaction produces #x# moles of carbon dioxide and #x# moles of water vapor, as given by the fact that you have a #1:1# mole ratio between the two products.

You can use the molar masses of the two products to express their respective mass in terms of #x#

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

#x color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = (18.015 * x)" g H"_2"O"#

You can now use equation #color(orange)("(*)")# to write

#overbrace(44.01 * x)^(color(blue)("mass of CO"_2)) color(red)(cancel(color(black)("g")))" "+" " overbrace(18.015 * x)^(color(blue)("mass of H"_2"O"))color(red)(cancel(color(black)("g"))) = 2.28color(red)(cancel(color(black)("g")))#

Solve this equation to find

#62.025 * x = 2.28 implies x= 2.28/62.025 = 0.03676#

Now, you know that the reaction produced #0.03676# moles of carbon dioxide and #0.03676# moles of water vapor.

Notice that you have a #color(darkgreen)(2):1# mole ratio between sodium bicarbonate and carbon dioxide. This tells you that the original mixture contained

#0.03676 color(red)(cancel(color(black)("moles CO"_2))) * (color(darkgreen)(2)color(white)(a)"moles NaHCO"_3)/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.07352 moles NaHCO"_3#

Use the molar mass of sodium bicarbonate to convert the number of moles to grams

#0.07352 color(red)(cancel(color(black)("moles NaHCO"_3))) * "84.007 g"/(1color(red)(cancel(color(black)("mole NaHCO"_3)))) = "6.176 g NaHCO"_3#

You now know that your mixture contained #"6.176 g"# of sodium bicarbonate and

#14.00 - 6.176 = "7.824 g Na"_2"CO"_3#

This means that the percent composition of sodium carbonate in the mixture was

#"% Na"_2"CO"_3 = (7.824 color(red)(cancel(color(black)("g"))))/(14.00color(red)(cancel(color(black)("g")))) * 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(55.9%)color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Now, let's tackle the limiting reactant part. Right from the start you should be able to tell that sodium carbonate cannot be the limiting reactant because it doesn't take part in the reaction.

Heating the mixture does not cause the sodium carbonate to decompose, so there's no reason to consider it as a limiting reactant because it's not even a reactant.

However, this thing to keep in mind here is that because you're dealing with a decomposition reaction, you can't really talk about a limiting reactant.

Your reaction involves the decomposition of sodium bicarbonate. You only have one reactant here, which means that the amount of product you get will always depend on the amount of one reactant.

My answer would be #-># the reaction doesn't have a limiting reactant because it only has one reactant, sodium bicarbonate. A reactant can't limit itself, so we can't talk about a limiting reactant here.