Question

What is the maximum mass of ${\text{Al}}$ (molar mass $27{\text{ g/mol}}$) that can be obtained from $20.4{\text{ g}}$ of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ (molar mass $102{\text{ g/mol}}$)?
A) $2.70{\text{ g}}$
B) $5.40{\text{ g}}$
C) $8.10{\text{ g}}$
D) $10.8{\text{ g}}$
E) $16.3{\text{ g}}$

Answer 1

To solve this we must first write the correct balanced chemical equation for the conversion of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ to ${\text{Al}}$. From the balanced chemical equation, we can know the number of moles of both ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and ${\text{Al}}$ and thus, we can calculate the mass of ${\text{Al}}$.

Complete solution:
We know that ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ when heated decomposes to form ${\text{Al}}$ and oxygen gas is released. The decomposition reaction of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is as follows:
${\text{2A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} \to 4{\text{Al}} + 3{{\text{O}}_2}$
From the balanced chemical equation we can see that two moles of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ decompose and produce four moles of ${\text{Al}}$. Thus,
${\text{2 mol A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 4{\text{ mol Al}}$
${\text{1 mol A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 2{\text{ mol Al}}$
We know that the number of moles of any substance is the ratio of its mass to molar mass. Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Rearrange the equation for the mass as follows:
${\text{Mass}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molar mass}}\left( {{\text{g/mol}}} \right)$
Calculate the mass of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ in ${\text{1 mol}}$ of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ using the equation as follows:
${\text{Mass of A}}{{\text{l}}_2}{{\text{O}}_3} = 1{\text{ mol}} \times 102{\text{ g/mol}}$
${\text{Mass of A}}{{\text{l}}_2}{{\text{O}}_3} = 102{\text{ g}}$
Thus, the mass of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ in ${\text{1 mol}}$ of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $102{\text{ g}}$.
Calculate the mass of ${\text{Al}}$ in ${\text{2 mol}}$ of ${\text{Al}}$ using the equation as follows:
${\text{Mass of Al}} = 2{\text{ mol}} \times 27{\text{ g/mol}}$
${\text{Mass of Al}} = 54{\text{ g}}$
Thus, the mass of ${\text{Al}}$ in ${\text{2 mol}}$ of ${\text{Al}}$ is $54{\text{ g}}$.
Thus, $102{\text{ g}}$ of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ produces $54{\text{ g}}$ of ${\text{Al}}$. Thus, the mass of ${\text{Al}}$ produced by $20.4{\text{ g}}$ of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is,
${\text{Mass of Al}} = 20.4{\text{ g A}}{{\text{l}}_2}{{\text{O}}_3} \times \dfrac{{54{\text{ g Al}}}}{{102{\text{ g A}}{{\text{l}}_2}{{\text{O}}_3}}}$
${\text{Mass of Al}} = 10.8{\text{ g}}$
Thus, the maximum mass of ${\text{Al}}$ that can be obtained from $20.4{\text{ g}}$ of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $10.8{\text{ g}}$.

Thus, the correct option is (D) $10.8{\text{ g}}$.

Note: Remember that the balanced chemical equation for the decomposition of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ must be written correctly. Incorrect or unbalanced chemical equations can lead to an incorrect number of moles which can lead to incorrect mass of the element.