Question

How much aluminium oxide and how much carbon are needed to prepare $454{\text{ g}}$ of aluminium by the balance equation?

Answer 1

Hint:To solve this first write the balanced chemical reaction for the preparation of aluminium from aluminium oxide and carbon. Then from the reaction stoichiometry calculate the mole ratio between aluminium oxide and aluminium and carbon and aluminium. From the mole ratio, calculate mass of aluminium oxide and carbon.

Complete solution:
We are given that aluminium i.e. ${\text{Al}}$ is prepared from aluminium oxide i.e. ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and carbon i.e. ${\text{C}}$. The balanced reaction for the preparation of aluminium from aluminium oxide and carbon is as follows:
${\text{2A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 3{\text{C}} \to 4{\text{Al}} + 3{\text{C}}{{\text{O}}_2}$
We are given $454{\text{ g}}$ of aluminium. Calculate the number of moles of aluminium in $454{\text{ g}}$ of aluminium as follows:
We know that the number of moles is the ratio of mass to the molar mass. Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $454{\text{ g}}$ for the mass of aluminium, $27{\text{ g/mol}}$ for the molar mass of aluminium. Thus,
${\text{Number of moles of Al}} = \dfrac{{454{\text{ g}}}}{{27{\text{ g/mol}}}}$
${\text{Number of moles of Al}} = 16.81{\text{ mol}}$
Thus, the number of moles of aluminium in $454{\text{ g}}$ of aluminium are $16.81{\text{ mol}}$.
From the balanced reaction stoichiometry, we can say that two moles of aluminium oxide correspond to four moles of aluminium. Thus,
${\text{2 mol A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 4{\text{ mol Al}}$
Thus, the number of moles of aluminium oxide needed to prepare $16.81{\text{ mol}}$ aluminium are,
${\text{Number of moles of A}}{{\text{l}}_2}{{\text{O}}_3} = 16.81{\text{ mol Al}} \times \dfrac{{2{\text{ mol A}}{{\text{l}}_2}{{\text{O}}_3}}}{{4{\text{ mol Al}}}}$
${\text{Number of moles of A}}{{\text{l}}_2}{{\text{O}}_3} = 8.405{\text{ mol}}$
Thus, the number of moles of aluminium oxide needed to prepare $19.73{\text{ mol}}$ aluminium are $8.405{\text{ mol}}$.
Calculate the mass of aluminium oxide in $8.405{\text{ mol}}$ of aluminium oxide as follows:
${\text{Mass}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molar mass}}\left( {{\text{g/mol}}} \right)$
Substitute $8.405{\text{ mol}}$ for the number of moles of aluminium oxide and $102{\text{ g/mol}}$ for the molar mass of aluminium oxide. Thus,
${\text{Mass of A}}{{\text{l}}_2}{{\text{O}}_3} = 8.405{\text{ mol}} \times 102{\text{ g/mol}}$
${\text{Mass of A}}{{\text{l}}_2}{{\text{O}}_3} = 857.31{\text{ g}}$
Thus, the amount of aluminium oxide needed to prepare $454{\text{ g}}$ of aluminium is $857.31{\text{ g}}$.
From the balanced reaction stoichiometry, we can say that three moles of carbon correspond to four moles of aluminium. Thus,
${\text{2 mol C}} = 4{\text{ mol Al}}$
Thus, the number of moles of carbon needed to prepare $16.81{\text{ mol}}$ aluminium are,
${\text{Number of moles of C}} = 16.81{\text{ mol Al}} \times \dfrac{{3{\text{ mol C}}}}{{4{\text{ mol Al}}}}$
${\text{Number of moles of C}} = 12.6075{\text{ mol}}$
Thus, the number of moles of carbon needed to prepare $19.73{\text{ mol}}$ aluminium are $12.6075{\text{ mol}}$.
Calculate the mass of carbon in $12.6075{\text{ mol}}$ of carbon as follows:
${\text{Mass}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molar mass}}\left( {{\text{g/mol}}} \right)$
Substitute $12.6075{\text{ mol}}$ for the number of moles of carbon and $12{\text{ g/mol}}$ for the molar mass of carbon. Thus,
${\text{Mass of C}} = 12.6075{\text{ mol}} \times 12{\text{ g/mol}}$
${\text{Mass of C}} = 151.29{\text{ g}}$
Thus, the amount of carbon needed to prepare $454{\text{ g}}$ of aluminium is $151.29{\text{ g}}$.

Thus, the amount of aluminium oxide and carbon needed to prepare $454{\text{ g}}$ of aluminium are $857.31{\text{ g}}$ and $151.29{\text{ g}}$ respectively.

Note:Remember that the balanced chemical equation for the given reaction 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.