Question

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"`?

Answer 1

Aluminium (Al)

Explanation:

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

Answer 2

For the reaction

`"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"`

`"Fe"_2"O"_3` is `1:1` with `"X"_2"O"_3`, so `"0.50001 mols"` of `"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` `"g"`

`=` `"26.982 g X"`

Since `"X"` is `2:1` with `"X"_2"O"_3`, `"1.00002 mols"` of `"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.