# The mixture of Na2CO3 and NaHCO3 weighs 220g. Heated precipitate weighs 182g. What is the percent composition?

Mar 13, 2016

The mixture is ${\text{53.2 % Na"_2"CO}}_{3}$ and ${\text{46.8 % NaHCO}}_{3}$.

#### Explanation:

When you heat a mixture of ${\text{Na"_2"CO}}_{3}$ and ${\text{NaHCO}}_{3}$, only the ${\text{NaHCO}}_{3}$ decomposes.

The equation for the decomposition is

"2NaHCO"_3("s") → "Na"_2"CO"_3("s") + "CO"_2("g") + "H"_2"O"("g")

The loss in mass is caused by the loss of $\text{H"_2"O}$ and ${\text{CO}}_{2}$ (equivalent to the loss of ${\text{H"_2"CO}}_{3}$).

The loss in mass is $\text{(220 – 182) g = 38 g}$.

This corresponds to

38 color(red)(cancel(color(black)("g H"_2"CO"_3))) × ("1 mol H"_2"CO"_3)/(62.02 color(red)(cancel(color(black)("g H"_2"CO"_3)))) = "0.613 mol H"_2"CO"_3

${\text{Moles of NaHCO"_3 = 0.613 color(red)(cancel(color(black)("mol H"_2"CO"_3))) × ("2 mol NaHCO"_3)/(1 color(red)(cancel(color(black)("mol H"_2"CO"_3)))) = "1.225 mol NaHCO}}_{3}$

and

${\text{Mass of NaHCO"_3 = 1.2225 color(red)(cancel(color(black)("mol NaHCO"_3))) × ("84.01 g NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaHCO"_3)))) = "102.9 g NaHCO}}_{3}$.

The loss of mass after heating corresponds to ${\text{102.9 g of original NaHCO}}_{3}$.

The rest of the original mixture must have been ${\text{Na"_2"CO}}_{3}$.

$\text{Mass of original Na"_2"CO"_3 = "220 g – 102.9 g" = "117.1 g}$

$\text{% Na"_2"CO"_3 = (117.1 color(red)(cancel(color(black)("g"))))/(220 color(red)(cancel(color(black)("g")))) × 100 % = "53.2 %}$

% "NaHCO"_3 = (102.9 color(red)(cancel(color(black)("g"))))/(220 color(red)(cancel(color(black)("g")))) × 100 % = 46.8 %