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"#?

2 Answers
Aug 18, 2017

Answer:

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.

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.