Chapter 10: Problem 146
Calculate the moles of aluminum ions present in 250.0 \(\mathrm{g}\) of aluminum oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\)
Short Answer
Expert verified
In 250.0 g of aluminum oxide (Al鈧侽鈧), there are 4.8992 moles of aluminum ions (Al鲁鈦) present.
Step by step solution
01
Find the molar mass of Al鈧侽鈧
To do this, we first need to find the molar masses of aluminum (Al) and oxygen (O) individually. The molar mass of aluminum is 26.98 g/mol, and that of oxygen is 16.00 g/mol.
Now we can calculate the molar mass of Al鈧侽鈧:
Molar mass of Al鈧侽鈧 = (2 脳 molar mass of Al) + (3 脳 molar mass of O)
Molar mass of Al鈧侽鈧 = (2 脳 26.98 g/mol) + (3 脳 16.00 g/mol)
Molar mass of Al鈧侽鈧 = 53.96 g/mol + 48.00 g/mol
Molar mass of Al鈧侽鈧 = 101.96 g/mol
02
Calculate the moles of Al鈧侽鈧 present in 250.0 g
To do this, we can use the given mass of Al鈧侽鈧 and its molar mass:
Moles of Al鈧侽鈧 = (mass of Al鈧侽鈧) / (molar mass of Al鈧侽鈧)
Moles of Al鈧侽鈧 = (250.0 g) / (101.96 g/mol)
Moles of Al鈧侽鈧 = 2.4496 mol
03
Determine the moles of aluminum ions (Al鲁鈦) in the given amount of Al鈧侽鈧
In one mole of Al鈧侽鈧, there are 2 moles of Al ions. To find the moles of Al ions present in 2.4496 moles of Al鈧侽鈧, use the below relationship:
Moles of Al鲁鈦 = (2 脳 moles of Al鈧侽鈧)
Moles of Al鲁鈦 = 2 脳 2.4496 mol
Moles of Al鲁鈦 = 4.8992 mol
Therefore, there are 4.8992 moles of aluminum ions present in 250.0 g of aluminum oxide.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass Calculation
To calculate the molar mass of a compound, you need to know the atomic masses of the constituent elements. This is essential for converting the mass of a substance to the number of moles, a fundamental concept in chemistry.
In the case of aluminum oxide \((\text{Al}_2\text{O}_3)\), you start by identifying the individual atomic masses:
This molar mass is used to relate the mass of the compound to the number of moles.
In the case of aluminum oxide \((\text{Al}_2\text{O}_3)\), you start by identifying the individual atomic masses:
- Aluminum (Al): 26.98 \(\text{g/mol}\)
- Oxygen (O): 16.00 \(\text{g/mol}\)
- For Al: \(2 \times 26.98 \text{ g/mol} = 53.96 \text{ g/mol}\)
- For O: \(3 \times 16.00 \text{ g/mol} = 48.00 \text{ g/mol}\)
This molar mass is used to relate the mass of the compound to the number of moles.
Aluminum Oxide
Aluminum oxide is a chemical compound with the formula \(\text{Al}_2\text{O}_3\). It's commonly found in several mineral forms, notably in bauxite, the principal ore of aluminum.
This compound is important in both industrial and manufacturing applications, including metallurgy and ceramics.
In aluminum oxide:
This compound is important in both industrial and manufacturing applications, including metallurgy and ceramics.
In aluminum oxide:
- There are 2 aluminum atoms.
- There are 3 oxygen atoms.
Moles Calculation
The concept of "moles" in chemistry refers to a standard method for expressing quantities of atoms, molecules, or ions. It's like a bridge between the atomic scale and the macroscopic scale we can measure.
A mole is much like a dozen鈥攊t simply means 6.022 x 10虏鲁 of something, which is also known as Avogadro鈥檚 number.
To find the number of moles in a given mass of a substance, you use the formula:
A mole is much like a dozen鈥攊t simply means 6.022 x 10虏鲁 of something, which is also known as Avogadro鈥檚 number.
To find the number of moles in a given mass of a substance, you use the formula:
- \( \text{Moles} = \frac{\text{Mass of substance}}{\text{Molar mass}} \)
Chemical Composition
Chemical composition refers to the elements that make up a compound and the ratio in which they are present. Understanding this is critical for calculating chemical quantities accurately and predicting the behavior of compounds in various conditions.
The composition of \(\text{Al}_2\text{O}_3\) breaks down into:
Knowing the chemical composition aids in numerous applications such as determining the necessary amounts of reactants in a chemical reaction or creating desired mixtures in compound synthesis.
The composition of \(\text{Al}_2\text{O}_3\) breaks down into:
- Two atoms of aluminum (Al),
- Three atoms of oxygen (O).
- Al contributes \(53.96 \text{ g/mol}\).
- O contributes \(48.00 \text{ g/mol}\).
Knowing the chemical composition aids in numerous applications such as determining the necessary amounts of reactants in a chemical reaction or creating desired mixtures in compound synthesis.
Stoichiometry
Stoichiometry is the branch of chemistry involved with calculating the quantities of reactants and products in chemical reactions. This concept applies the law of conservation of mass, which states that mass in a chemical reaction is neither created nor destroyed.
By using balanced chemical equations, stoichiometry allows you to predict the amounts of substances consumed and produced.
In terms of aluminum oxide, stoichiometry comes into play when you understand that one mole of \(\text{Al}_2\text{O}_3\) contains two moles of aluminum ions (Al鲁鈦). Thus, if you have 2.4496 moles of \(\text{Al}_2\text{O}_3\), it translates to 4.8992 moles of aluminum ions, as calculated by the stoichiometric coefficient.
Understanding stoichiometry is key to scaling laboratory reactions to industrial applications, ensuring that complex reactions can be managed and optimized for various chemical processes.
By using balanced chemical equations, stoichiometry allows you to predict the amounts of substances consumed and produced.
In terms of aluminum oxide, stoichiometry comes into play when you understand that one mole of \(\text{Al}_2\text{O}_3\) contains two moles of aluminum ions (Al鲁鈦). Thus, if you have 2.4496 moles of \(\text{Al}_2\text{O}_3\), it translates to 4.8992 moles of aluminum ions, as calculated by the stoichiometric coefficient.
Understanding stoichiometry is key to scaling laboratory reactions to industrial applications, ensuring that complex reactions can be managed and optimized for various chemical processes.