# Question 22a3c

Jun 17, 2017

${\text{63% NaHCO}}_{3}$

$\text{37% NaCl}$

#### Explanation:

The idea here is that the sodium bicarbonate will undergo thermal decomposition to produce sodium carbonate, water, and carbon dioxide as described by the following balanced chemical equation

$\textcolor{b l u e}{2} {\text{NaHCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "Na"_ 2"CO"_ (3(s)) + "CO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

On the other hand, the sodium chloride will not undergo thermal decomposition, so you know for a fact that all the mass of carbon dioxide produced by heating the mixture came from the sodium bicarbonate.

Use the molar mass of carbon dioxide to convert the mass to moles

0.66 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.0150 moles CO"_2

As you can see, $1$ mole of carbon dioxide is produced by the reaction when $\textcolor{b l u e}{2}$ moles of sodium bicarbonate undergo decomposition, so you can say that the mixture contained

0.0150 color(red)(cancel(color(black)("moles CO"_2))) * (color(blue)(2)color(white)(.)"moles NaHCO"_3)/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.0300 moles NaHCO"_3

To convert this to grams, use the molar mass of sodium bicarbonate

0.0300 color(red)(cancel(color(black)("moles NaHCO"_3))) * "84.007 g"/(1color(red)(cancel(color(black)("mole NaHCO"_3)))) = "2.52 g"

This means that percent composition of the mixture was

"% NaHCO"_3 = (2.52 color(red)(cancel(color(black)("g"))))/(4color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(63%)))

"% NaCl" = 100% - "% NaHCO"_3 = color(darkgreen)(ul(color(black)(37%)))#

I'll leave the answers rounded to two sig figs, but keep in mind that you only have one significant figuer for the mass of the mixture.