Question

For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s) + aluminum (s) aluminum oxide (s) + iron (s) What is the maximum amount of aluminum oxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Iron (III) oxide is a limiting reagent i.e $Fe_2O_3$.

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

$Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)$

Moles of  iron(III) oxide : $\frac{53.7 g}{160 g/mol}=0.3356 mol$

Moles of aluminium : $\frac{22.8 g}{27 g/mol}=0.8444 mol$

According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.

Then 0.3356 moles of iron(III) oxide will react with:

$\frac{2}{1}\times 0.3356 mol=0.6712 mol$ of aluminum.

As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e $Fe_2O_3$ and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

$\frac{1}{1}\times 0.3356 mol=0.3356 mol$ of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete