# For the reaction "Fe"_2"O"_3 + 2"X" -> 2"Fe" + "X"_2"O"_3, if "79.847 g" "Fe"_2"O"_3 reacts with excess "X" (FW = "159.691 g/mol" for the iron(III) oxide) to yield "50.982 g X"_2"O"_3 and "55.847 g Fe", what is the identity of "X"?

Aug 18, 2017

Aluminium (Al)

#### Explanation:

There is a simple concept.
According to law of mass conservation,
$\left[2 X\right] = \left(55.847 + 50.982\right) - 79.847 = 26.982$
Here, $2 F e$ has a mass same as its atomic weight so as $2 X$ has a mass same as its atomic weight which is $26.982$. This is the atomic weight of Aluminium.

Aug 19, 2017

For the reaction

${\text{Fe"_2"O"_3 + 2"X" -> 2"Fe" + "X"_2"O}}_{3}$,

convert the mass of iron(III) oxide to mols.

79.847 cancel("g Fe"_2"O"_3) xx ("1 mol")/(159.691 cancel("g Fe"_2"O"_3)) = "0.50001 mols"

${\text{Fe"_2"O}}_{3}$ is $1 : 1$ with ${\text{X"_2"O}}_{3}$, so $\text{0.50001 mols}$ of ${\text{X"_2"O}}_{3}$ was produced. By conservation of mass,

m_"X" = m_("X"_2"O"_3) + m_"Fe" - m_("Fe"_2"O"_3)

$= 50.982 + 55.847 - 79.847$ $\text{g}$

$=$ $\text{26.982 g X}$

Since $\text{X}$ is $2 : 1$ with ${\text{X"_2"O}}_{3}$, $\text{1.00002 mols}$ of $\text{X}$ was produced. So, its molar mass is:

"26.982 g X"/"1.00002 mols" = ul(26.981_5color(white)(.)"g/mol")

which is essentially the exact molar mass of aluminium (aluminum) atom.