Question

In the equation, $F{e_2}{O_3} + 2Al \to A{l_2}{O_3} + 2Fe$ if $52.6g$ of aluminium are used with excess iron oxide, how many grams of $F{e_2}{O_3}$ will be converted to aluminium oxide?

Answer 1

From the balanced chemical equation, the moles of one reactant can be determined from the moles of another reactant. The amount of aluminium is given as $52.6g$ . The number of moles of aluminium is double to the moles of $F{e_2}{O_3}$ . By determining the amount of one gram of aluminium reacted, the grams of $F{e_2}{O_3}$ will be converted to aluminium oxide.

Complete answer:
Given that ferric oxide or iron oxide reacts with aluminium metal liberates aluminium oxide and iron metal. This reaction can be considered a single displacement reaction.
Given reaction is $F{e_2}{O_3} + 2Al \to A{l_2}{O_3} + 2Fe$ ,
Given that $52.6g$ of aluminium is reacted with excess iron oxide.
From the balanced chemical equation, it was clear that the number of moles of aluminum is double the number of moles of iron oxide.
Thus, $54g$ of $Al$ reacts with $160g$ of $F{e_2}{O_3}$ as the molar mass of $Al$ is $23amu$ and molar mass of $F{e_2}{O_3}$ is $160amu$
By this, $1g$ of $Al$ reacts with $\dfrac{{160}}{{54}}g$ of $F{e_2}{O_3}$
As the given mass of $Al$ is $52.6g$ , it reacts with $\dfrac{{160 \times 52.6}}{{54}}g$ of $F{e_2}{O_3}$
By simplification, it is equal to $155.85g$ of $F{e_2}{O_3}$
Thus, $52.6g$ of aluminium is used with excess iron oxide, $155.85$ grams of $F{e_2}{O_3}$ will be converted to aluminium oxide.

Note:
Based on the balanced chemical equation only, the moles of reactants and products were calculated. The molar mass of aluminium metal must be multiplied by two, as two moles of aluminium were reacted with iron oxide to form aluminium oxide.