Question

In a reaction, 34.0 g of chromium(III) oxide reacts with 12.1 g of aluminum to produce chromium and aluminum oxide. If 23.3 g of chromium is produced, what mass of aluminum oxide is produced?

Short answer

The mass of aluminum oxide produced in the reaction is approximately \(22.85\ g\).

Step-by-step solution

Find the molar masses of the involved compounds

First, we need to find the molar masses of Cr2O3, Al, Cr, and Al2O3. To do this, we will use their atomic masses from the periodic table: - Atomic mass of Cr: 51.996 g/mol - Atomic mass of Al: 26.982 g/mol - Atomic mass of O: 16.00 g/mol Now, we can find the molar masses of chromium(III) oxide and aluminum oxide: - Molar mass of Cr2O3: \(2(51.996) + 3(16.00) = 151.99\) g/mol - Molar mass of Al2O3: \(2(26.982) + 3(16.00) = 101.96\) g/mol

Convert the given masses to moles

Next, we will convert the given masses of chromium(III) oxide, aluminum, and chromium to moles using their molar masses: - Moles of Cr2O3: \(\frac{34.0\ g}{151.99\ \text{g/mol}} = 0.2238\) mol - Moles of Al: \(\frac{12.1\ g}{26.982\ \text{g/mol}} = 0.4485\) mol - Moles of Cr: \(\frac{23.3\ g}{51.996\ \text{g/mol}} = 0.4477\) mol

Use stoichiometry to determine moles of aluminum oxide produced

The balanced chemical reaction between chromium(III) oxide and aluminum is: \(Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3\) Using stoichiometry, for 2 moles of aluminum consumed, 1 mole of aluminum oxide is produced. We have already converted the mass of aluminum to moles, so we can use that to find the moles of aluminum oxide produced: Moles of Al2O3: \(\frac{1\ mol\ Al_2O_3}{2\ mol\ Al} × 0.4485\ mol\ Al = 0.2243 \ mol\ Al_2O_3\)

Convert moles of aluminum oxide to mass

Finally, we can convert the moles of aluminum oxide back to its mass using its molar mass: Mass of Al2O3: \(0.2243\ mol\ Al_2O_3 × 101.96\ \text{g/mol} = 22.85\ g\) So, the mass of aluminum oxide produced is approximately \(22.85\ g\).

Key concepts

Molar Mass Calculations

Calculating the molar mass is a fundamental step in any stoichiometry problem. Molar mass, expressed in grams per mole (g/mol), is crucial for converting between the mass of a substance and the number of moles. This is essential because chemical reactions happen in terms of moles rather than mass.
To find the molar mass, you need the atomic masses of the elements involved, which are readily available on the periodic table. For chromium(III) oxide,

• each chromium (Cr) atom has an atomic mass of about 52 g/mol, and each oxide ion (O) is about 16 g/mol. So, for Cr₂O₃, the calculation is: 2 Cr atoms = 2(51.996 g/mol) and 3 O atoms = 3(16.00 g/mol), totalling 151.99 g/mol.

Similarly, for aluminum oxide (Al₂O₃),

• each aluminum (Al) atom has an atomic mass of about 27 g/mol. Therefore, 2 Al atoms = 2(26.982 g/mol) and the same 3 O atoms = 3(16 g/mol), totalling 101.96 g/mol.

These calculated molar masses serve as conversion factors so you can work with the substances in moles, simplifying calculations and ensuring correct proportions in the reaction.

Balanced Chemical Equation

A balanced chemical equation is key for understanding and solving stoichiometry problems. It represents the law of conservation of mass, indicating that matter cannot be created or destroyed. This means that in a chemical reaction, the number of atoms of each element must be the same on both sides of the equation.
For the reaction between chromium(III) oxide and aluminum, the balanced equation is:

Cr₂O₃ + 2Al → 2Cr + Al₂O₃

This equation tells us:

• The coefficients show the ratio of reactants and products involved – 1 mole of Cr₂O₃ reacts with 2 moles of Al to produce 2 moles of Cr and 1 mole of Al₂O₃.
• It serves as a guide for how much of each substance you need to start with and how much will be produced. This ensures calculations are accurate and reflect the actual reaction.

Understanding and accurately writing a balanced equation helps dictate how substances should be converted and used in calculations.

Conversion of Mass to Moles

Converting mass to moles is a critical step in solving stoichiometry problems, as reactions are based on mole proportions rather than mass. The conversion uses the molar mass of a substance as the conversion factor.
To convert mass to moles, you apply the formula: ewline \[\text{Number of Moles} = \frac{\text{Mass of the substance}}{\text{Molar Mass}}\]This formula allows the reaction to be considered at the atomic level and provides a clearer view of the reactant and product ratios.
For example, in the given problem:

• Mass of Cr₂O₃ = 34.0 g -> Moles of Cr₂O₃ = \(\frac{34.0\ g}{151.99\ \text{g/mol}} = 0.2238\ \text{mol}\).
• Mass of Al = 12.1 g -> Moles of Al = \(\frac{12.1\ g}{26.982\ \text{g/mol}} = 0.4485\ \text{mol}\).

Once you have the moles, you can use them to find out how many moles of product are formed using the balanced chemical equation. Understanding this allows you to track the amount of each reactant and product, ensuring consistency in experimental and theoretical yields.