Chemistry, 10.10.2019 02:00 thetudent41

Calculate the number of moles of al2o3 that are produced when 0.60 mol of fe is produced in the following reaction: 2al(s)+3feo(s) = 3fe(s)+al2o3(s)

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2 Al(s) + 3 FeO(s) = 3 Fe(s)+Al₂O₃(s)

3 moles Fe > 1 mole Al₂O₃
0.60 moles Fe > ?

moles Al₂O3 = 0.60 x 1 / 3

moles Al₂O₃ = 0.60 / 3

= 0.2 moles of Al₂O₃

hope this helps!

Answer from: manco6

The number of mole of ${\text{A}}{{\text{l}}_2}{{\text{O}}_{\text{3}}}$ produced is $\boxed{{\text{0}}{\text{.20 mol}}}$ .

Further explanation:

A balanced chemical reaction is a chemical reaction that has equal number of atoms of the different elements in the reactant as well as on the product side. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

Stoichiometry of a reaction is used to determine the number of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a compound when the moles of another compound in the reaction are given.

The general reaction is as follows:

${\text{A}}+2{\text{B}}\to3{\text{C}}$

Here,

A and B are reactants.

C is the product.

One mole of Aand two moles of B react to form three moles of C. The stoichiometric ratio between A and B is 1:2, between A and C is 1:3 and between B and C is 2:3.

The given chemical reaction is as follows:

${\text{2\;Al}}\left({\text{s}}\right)\;+\;{\text{3\;FeO}}\left({\text{s}}\right)\to {\text{3\;Fe}}\left({\text{s}}\right)+{\text{A}}{{\text{l}}_2}{{\text{O}}_{\text{3}}}\left({\text{s}}\right)$

According to the stoichiometry of the above-balanced chemical reaction, 2 moles of Al react with 3 moles of FeO to produce 3 moles of Fe and 1 mole of ${\text{A}}{{\text{l}}_2}{{\text{O}}_{\text{3}}}$ .

Therefore, the ratio of the moles of Fe and ${\text{A}}{{\text{l}}_2}{{\text{O}}_{\text{3}}}$ will be 3:1.

The formula to determine the number of moles of ${\text{A}}{{\text{l}}_2}{{\text{O}}_{\text{3}}}$ is written as follows:

${\text{Number of moles }}\left({{\text{A}}{{\text{l}}_2}{{\text{O}}_{\text{3}}}} \right)=\frac{1}{3}{\text{(number of moles of Fe}})$ …… (1)

Substitute 0.60 mol for number of moles of Fe in equation (1).

$\begin{aligned}{\text{Number of moles}}\left({{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}}\right)&=\frac{{\text{1}}}{{\text{3}}}\left({{\text{0}}{\text{.60 mol}}}\right)\\&={\text{0}}{\text{.20 mol}}\\\end{aligned}$

Hence, 0.20 mol of ${\mathbf{A}}{{\mathbf{l}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}$ is produced when 0.60 mol of Fe is produced in the reaction.

Learn more:

1. Write the equation to show how cation acts as an acid link.

2. Which are the characteristics of mixtures link?

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: Balanced chemical reaction, stoichiometry, reactants, products, number of moles, Fe, Al2O3, FeO and Al.

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Calculate the number of moles of al2o3 that are produced when 0.60 mol of fe is produced in the foll...

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