Aluminum metal reacts with iron (III) oxide to produce aluminum oxide and iron metal. Answers the following three questions
note: you will need to write a balanced equation to answer them
Al: 26.9 amu Fe: 55.8 amu O: 16 amu
1 How many moles of iron (III) oxide are required to react completely with 41 g of aluminum metal?
2 How many grams of aluminum metal are requred to produce 7.0 g of iron metal?
To answer these questions, let's start by writing the balanced equation for the reaction:
2 Al + Fe2O3 -> Al2O3 + 2 Fe
1. To determine the number of moles of iron (III) oxide required, we need to use the molar mass of Fe2O3.
Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O) = (2 x 55.8) + (3 x 16) = 111.6 + 48 = 159.6 g/mol
Now, we can determine the number of moles of Fe2O3 by dividing the given mass of aluminum metal by its molar mass:
Moles of Fe2O3 = Mass of Al metal (g) / Molar mass of Fe2O3 (g/mol)
Moles of Fe2O3 = 41 g / 159.6 g/mol ≈ 0.257 mol
Therefore, approximately 0.257 moles of iron (III) oxide are required to react completely with 41 g of aluminum metal.
2. To determine the mass of aluminum metal required to produce 7.0 g of iron metal, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that 2 moles of aluminum react to produce 2 moles of iron.
Molar mass of Al = 26.9 g/mol
Molar mass of Fe = 55.8 g/mol
Now, we can calculate the mass of aluminum required using the following proportion:
(2 moles Al / 2 moles Fe) = (X grams Al / 7.0 g Fe)
Cross-multiplying and solving for X, we get:
X grams Al = (2 moles Al / 2 moles Fe) x (7.0 g Fe)
X grams Al = (2/2) x (7.0 g) = 7.0 g
Therefore, 7.0 grams of aluminum metal are required to produce 7.0 grams of iron metal.
To answer these questions, we need to first write a balanced chemical equation for the reaction between aluminum metal and iron (III) oxide. The balanced equation is:
2Al + Fe2O3 -> Al2O3 + 2Fe
1. How many moles of iron (III) oxide are required to react completely with 41 g of aluminum metal?
In this question, we are given the mass of aluminum metal, and we need to find the number of moles of iron (III) oxide required for the reaction.
First, we need to convert the given mass of aluminum to moles. Aluminum has a molar mass of 26.9 g/mol.
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 41 g / 26.9 g/mol
Now, we use the balanced equation to find the ratio between moles of aluminum and moles of iron (III) oxide. From the equation, we see that 2 moles of Al reacts with 1 mole of Fe2O3.
Moles of Fe2O3 = (Moles of Al) / 2
Substituting the value we calculated earlier:
Moles of Fe2O3 = (41 g / 26.9 g/mol) / 2
Calculate this expression to find the number of moles of iron (III) oxide required for the reaction.
2. How many grams of aluminum metal are required to produce 7.0 g of iron metal?
In this question, we are given the mass of iron metal and asked to find the mass of aluminum required for the reaction.
First, we need to convert the given mass of iron to moles. Iron has a molar mass of 55.8 g/mol.
Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 7.0 g / 55.8 g/mol
Now, using the balanced equation, we can find the ratio between moles of aluminum and moles of iron. From the equation, we see that 2 moles of Al reacts with 2 moles of Fe.
Moles of Al = (Moles of Fe) / 2
Substituting the value we calculated earlier:
Moles of Al = (7.0 g / 55.8 g/mol) / 2
To find the mass of aluminum, we multiply the moles of aluminum by its molar mass:
Mass of Al = Moles of Al * Molar mass of Al
Calculate this expression to find the mass of aluminum required for the reaction.