Chemistry, 01.07.2019 13:00 amycressey1970

The following data was collected when a reaction was performed experimentally in the laboratory. fe2o3 al al2o3 fe starting amount in reaction 3 moles 5 moles ? ? determine the maximum amount of fe that was produced during the experiment. explain how you determined this amount.

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Answer from: clairesmom

The maximum amount of Fe produced will be 279.225 grams.

Explanation: The reaction follows:

$Fe_2O_3+2Al\rightarrow Al_2O_3+2Fe$

Given moles of Fe_2O_3 = 3 moles

Given moles of Al = 5 moles

By Stoichiometry of the reaction, we get

1 mole of Fe_2O_3 reacts with 2 moles of Al, so

3 moles of Fe_2O_3 will react with $(2\times 3)=6moles$ of Al

As the given moles of Al are less than the required amount, so it is considered as the limiting reagent as it limits the formation of product.

According to the reaction,

2 moles of Al produces 2 moles of Fe. Hence,

5 moles of Al will produce = $\frac{2}{2}\times 5=5moles$ of Fe.

To calculate the amount of Fe produces, we use the formula:

$Moles=\frac{\text{Given mass}}{\text{molar mass}}$

Moles of Fe = 4 moles

Molar mass of Fe = 55.845 g/mol

Putting the values in above equation, we get

$5=\frac{\text{Given mass}}{55.845g/mol}$

Mass of Fe produced = 279.225 grams

Answer from: suealex6578

$\boxed{279.225{\text{ g}}}$ of Fe is produced during the given experiment.

Further explanation:

The species or reagent in a chemical reaction that governs the amount of product is called the limiting reagent. It is completely consumed in the reaction.

Stoichiometry helps in determining the amount of species present in any reaction by establishing the relationship between reactants and products.

The balanced chemical reaction is as follows:

$\text{Fe}_2\text{O}_3+2\text{Al}\rightarrow\text{Al}_2\text{O}_3+2\text{Fe}$

The reaction stoichiometry states that one mole of $\text{Fe}_2\text{O}_3$ reacts with two moles of Al to produce one mole of $\text{Al}_2\text{O}_3$ and two moles of Fe.

We are provided with 3 moles of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$ and 5 moles of Al. Therefore the amount of Fe produced by 3 moles of ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of Fe}} &= \left( {3{\text{ mol F}}{{\text{e}}_2}{{\text{O}}_3}} \right)\left( {\frac{{2{\text{ mol Fe}}}}{{1{\text{ mol F}}{{\text{e}}_2}{{\text{O}}_3}}}} \right)\\&= 6{\text{ mol Fe}}\\\end{aligned}$

Similarly, the amount of Fe produced by 5 moles of Al can be calculated as follows:

$\begin{aligned}{\text{Amount of Fe}}&= \left( {5{\text{ mol Al}}} \right)\left( {\frac{{2{\text{ mol Fe}}}}{{2{\text{ mol Al}}}}} \right)\\&= 5{\text{ mol Fe}}\\\end{aligned}$

Since Al produces less amount of Fe than that formed by ${\text{F}}{{\text{e}}_2}{{\text{O}}_3}$ , Al is a limiting reagent. It will decide the amount of Fe produced during the reaction. Therefore 5 moles of Fe moles formed in this reaction.

The formula to calculate the moles of Fe is as follows:

${\text{Moles of Fe}} = \dfrac{{{\text{Mass of Fe}}}}{{{\text{Molar mass of Fe}}}}$ ...... (1)

Rearrange equation (1) to calculate the mass of Fe.

${\text{Mass of Fe}} = \left( {{\text{Moles of Fe}}} \right)\left( {{\text{Molar mass of Fe}}} \right)$ ...... (2)

Substitute 5 mol for the moles of Fe and 55.845 g/mol for the molar mass of Fe in equation (2).

$\begin{aligned}{\text{Mass of Fe}} &= \left( {{\text{5 mol}}} \right)\left( {{\text{55}}{\text{.845 g/mol}}} \right)\\&= 279.225{\text{ g}}\\\end{aligned}$