Question

Manganese metal and aluminum oxide are produced by the reaction of manganese(IV) oxide and aluminum metal. Calculate the mass of aluminum that is necessary to yield \(1.00 \mathrm{~kg}\) of manganese metal.

Short answer

655.5 g of aluminum is needed.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction between manganese(IV) oxide and aluminum can be written as:\[ \text{3 MnO}_2 + 4 \text{Al} \rightarrow \text{3 Mn} + 2 \text{Al}_2\text{O}_3 \]This equation shows that 4 moles of aluminum react with 3 moles of manganese(IV) oxide to produce 3 moles of manganese and 2 moles of aluminum oxide.

Determine Molar Masses

Calculate the molar masses of the substances involved: - Molar mass of Mn: 54.94 g/mol - Molar mass of Al: 26.98 g/mol These will be used to convert between mass and moles.

Convert Given Mass to Moles

We are given that 1.00 kg of manganese is produced. Convert this mass to moles using the molar mass of Mn:\[ \text{mass of Mn} = 1000 \text{ g} \]\[ \text{moles of Mn} = \frac{1000 \text{ g}}{54.94 \text{ g/mol}} \approx 18.21 \text{ mol} \]

Use Stoichiometry to Find Moles of Aluminum

The balanced equation shows that 3 moles of Mn are produced per 4 moles of Al. So, calculate the moles of Al required:\[ \frac{4 \text{ moles Al}}{3 \text{ moles Mn}} \times 18.21 \text{ mol Mn} = 24.28 \text{ mol Al} \]

Convert Moles of Aluminum to Mass

Now, convert the moles of Al to grams using its molar mass:\[ \text{mass of Al} = 24.28 \text{ mol} \times 26.98 \text{ g/mol} \approx 655.5 \text{ g} \]Therefore, 655.5 g of aluminum is needed to produce 1.00 kg of manganese.

Key concepts

Chemical Reactions

In chemistry, a chemical reaction describes the process where substances, known as reactants, transform into different substances, called products. This transformation involves the breaking of bonds in reactants and the formation of new bonds in products. Chemical reactions are essential for understanding how various substances interact and change.
In the given exercise, the reaction involves manganese(IV) oxide (\(\text{MnO}_2\)) and aluminum (\(\text{Al}\)) reacting to form manganese metal (\(\text{Mn}\)) and aluminum oxide (\(\text{Al}_2\text{O}_3\)). This specific type of reaction is a redox reaction, where one element is reduced and another is oxidized. Understanding this fundamental concept is crucial for determining the quantities of substances involved in the reaction, which is exactly what stoichiometry helps us achieve.

Molar Mass

The molar mass of a substance is a measure of the mass of one mole of that substance, typically expressed in grams per mole (\(\text{g/mol}\)). It is a crucial concept that allows chemists to convert between the mass of a substance and the number of moles present.
For example, in the exercise, the molar mass of manganese (Mn) is 54.94 g/mol and for aluminum (Al) is 26.98 g/mol. These values are critical for converting the mass of substances to and from moles, facilitating stoichiometric calculations. Calculating molar masses involves summing the atomic masses of all atoms present in a compound's formula. Mastery of this concept enables the solving of real-world chemistry problems, like determining the precise amount of reactant needed to yield a desired quantity of product.

Balanced Chemical Equation

A balanced chemical equation accurately represents the quantities of reactants and products in a chemical reaction. Each type of atom must occur in equal quantities on both sides of the equation to adhere to the law of conservation of mass.
In the example equation:\[\text{3 MnO}_2 + 4 \text{Al} \rightarrow \text{3 Mn} + 2 \text{Al}_2\text{O}_3\]This tells us 3 moles of manganese(IV) oxide react with 4 moles of aluminum to produce 3 moles of manganese and 2 moles of aluminum oxide. Balancing chemical equations is a fundamental step in stoichiometry: it ensures accurate calculations so chemists know precisely how much of each substance is needed or produced. Learning to balance equations effectively can greatly enhance one's understanding and capability in performing chemical calculations.