# Question 4cf9a

Jun 12, 2017

WARNING! Long answer! The mixture is 57 % Mohr's salt and 43 % ammonium sulfate. The mass of ${\text{Fe"_2"O}}_{3}$ obtained would be 23 mg.

#### Explanation:

The formula for Mohr's salt is ("NH"_4)_2"Fe"("SO"_4)_2·6"H"_2"O".

To avoid having to write complicated formulas, let's use $\text{M}$ to represent Mohr's salt and $\text{A}$ to represent ammonium sulfate.

The equations for the reaction with ${\text{BaCl}}_{2}$ are

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 123.14 \textcolor{w h i t e}{m m m m m m m l l} 233.39$
color(white)(mm)("NH"_4)_2"SO"_4 + "BaCl"_2 → "BaSO"_4 + "2NH"_4"Cl"

and

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m m m} 392.14 \textcolor{w h i t e}{m m m m m m m m m l l} 233.39$
color(white)(mm)("NH"_4)_2"Fe"("SO"_4)_2·6"H"_2"O" + "2BaCl"_2 → "2BaSO"_4 + "4NH"_4"Cl" + "FeCl"_2 + 6"H"_2"O"

Calculate the mass of ${\text{BaSO}}_{4}$ from $\text{A}$

Let $x$ be the mass of $\text{A}$. Then $0.5 - x$ is the mass of $\text{M}$.

${\text{Mass of BaSO"_4 = x color(red)(cancel(color(black)("g A"))) × (1 color(red)(cancel(color(black)("mol A"))))/(123.14 color(red)(cancel(color(black)("g A")))) × (1 color(red)(cancel(color(black)("mol BaSO"_4))))/(1 "mol A") × ("233.99 g BaSO"_4)/(1 color(red)(cancel(color(black)("mol BaSO"_4)))) = 1.9002 x color(white)(l)"g BaSO}}_{4}$

Calculate the mass of ${\text{BaSO}}_{4}$ from $\text{M}$

${\text{Mass of BaSO}}_{4}$

= (0.5 - x) color(red)(cancel(color(black)("g M"))) × (1 color(red)(cancel(color(black)("mol M"))))/(392.14 color(red)(cancel(color(black)("g M")))) × (2 color(red)(cancel(color(black)("mol BaSO"_4))))/(1 "mol M") × ("233.99 g BaSO"_4)/(1 color(red)(cancel(color(black)("mol BaSO"_4)))) = 1.1934(0.5-x) color(white)(l)"g BaSO"_4 = ("0.5967 - "1.1934x) "g BaSO"_4

Calculate the masses of $\text{A}$ and $\text{M}$

The total mass of ${\text{BaSO}}_{4}$ is 0.75 g.

$1.9002 x + \text{0.5967 - 1.1934} x = 0.75$

$0.7068 x = \text{0.75 - 0.5967} = 0.153$

$x = \frac{0.153}{0.7068} = 0.22$

The mass of $\text{A}$ is 0.22 g.

The total mass of $\text{A + M}$ is 0.5 g.

∴ The mass of $\text{M}$ is (0.5 - 0.22) g = 0.28 g

Calculate the % composition of the mixture

"% A" = (0.22 color(red)(cancel(color(black)("g"))))/(0.5 color(red)(cancel(color(black)("g")))) × 100 % = 43 % 

$\text{% M" = "(100 - 43) %" = 57 %}$

Calculate the mass of ${\text{Fe"_2"O}}_{3}$

The ${\text{Fe"_2"O}}_{3}$ comes from the Mohr's salt.

(a) Calculate the mass of $\text{M}$ in the sample

$\text{Mass of M in sample" = 0.2 color(red)(cancel(color(black)("g sample"))) × "57 g M"/(100 color(red)(cancel(color(black)("g Sample")))) = "0.114 g M}$

(b) Write the partial equation for the pyrolysis reaction

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m l l} 392.14 \textcolor{w h i t e}{m m m m m m m} 159.69$
color(white)(mm)2("NH"_4)_2"Fe"("SO"_4)_2·6"H"_2"O" → "Fe"_2"O"_3#

(c) Calculate the mass of ${\text{Fe"_2"O}}_{3}$

${\text{Mass of Fe"_2"O"_3 = 0.114 color(red)(cancel(color(black)("g M"))) × (1 color(red)(cancel(color(black)("mol M"))))/(392.14 color(red)(cancel(color(black)("g M")))) × (1 color(red)(cancel(color(black)("mol Fe"_2"O"_3))))/(2 color(red)(cancel(color(black)("mol M")))) × ("159.69 g Fe"_2"O"_3)/(1 color(red)(cancel(color(black)("mol Fe"_2"O"_3)))) = "0.023 g Fe"_2"O"_3 = "23 mg Fe"_2"O}}_{3}$