For the reaction ${Fe}_{2} O_{3} + 2 X \to 2 Fe + X_{2} O_{3}$ $"Fe"_2"O"_3 + 2"X" -> 2"Fe" + "X"_2"O"_3$ , if $79.847 g$ $"79.847 g"$ ${Fe}_{2} O_{3}$ $"Fe"_2"O"_3$ reacts with excess $X$ $"X"$ ( $F W = 159.691 g/mol$ $FW = "159.691 g/mol"$ for the iron(III) oxide) to yield ${50.982 g X}_{2} O_{3}$ $"50.982 g X"_2"O"_3$ and $55.847 g Fe$ $"55.847 g Fe"$ , what is the identity of $X$ $"X"$ ?

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For the reaction ${Fe}_{2} O_{3} + 2 X \to 2 Fe + X_{2} O_{3}$ $"Fe"_2"O"_3 + 2"X" -> 2"Fe" + "X"_2"O"_3$ , if $79.847 g$ $"79.847 g"$ ${Fe}_{2} O_{3}$ $"Fe"_2"O"_3$ reacts with excess $X$ $"X"$ ( $F W = 159.691 g/mol$ $FW = "159.691 g/mol"$ for the iron(III) oxide) to yield ${50.982 g X}_{2} O_{3}$ $"50.982 g X"_2"O"_3$ and $55.847 g Fe$ $"55.847 g Fe"$ , what is the identity of $X$ $"X"$ ?

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Answer 1 · 2017-08-18T15:22:23

Aluminium (Al)

Explanation:

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

Answer 2 · 2017-08-19T01:53:26

For the reaction

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

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.

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