Question

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

Short answer

The mass of aluminum oxide produced in the reaction is approximately \(34.2 \mathrm{~g}\).

Step-by-step solution

Write the balanced chemical equation for the reaction.

The given reaction is between chromium(III) oxide (Cr2O3) and aluminum (Al) to form chromium (Cr) and aluminum oxide (Al2O3). The balanced chemical equation for the given reaction is: \(2\,Cr_{2}O_{3} + 3\,Al \rightarrow 4\,Cr + 3\,Al_{2}O_{3}\)

Calculate the molar mass of each reactant and product.

The next step is to calculate the molar mass of each substance involved in the reaction. Use the periodic table to find the atomic mass of each element. Molar mass of \(Cr_2O_3\): (2 x 51.996) + (3 x 16.000) = 103.992 + 48.000 = 151.992 g/mol Molar mass of \(Al\): 26.981 g/mol Molar mass of \(Cr\): 51.996 g/mol Molar mass of \(Al_2O_3\): (2 x 26.981) + (3 x 16.000) = 53.962 + 48.000 = 101.961 g/mol

Convert grams of given substances to moles.

Now, we need to convert the given mass of chromium oxide, aluminum, and chromium produced into moles. Moles of \(Cr_2O_3\) = (34.0 g) / (151.992 g/mol) = 0.2237 mol Moles of \(Al\) = (12.1 g) / (26.981 g/mol) = 0.4486 mol Moles of \(Cr\) = (23.3 g) / (51.996 g/mol) = 0.4481 mol

Identify limiting reactant and calculate the moles of aluminum oxide formed.

To find the limiting reactant, we can compare the mole ratios: Mole ratio of chromium oxide to aluminum = [Moles of \(Cr_2O_3\)] / 2 : [Moles of \(Al\)] / 3 0.2237 / 2 : 0.4486 / 3 0.1119 : 0.1495 Since 0.1119 < 0.1495, chromium oxide is the limiting reactant. Now, we can use the stoichiometry from the balanced chemical equation to find the moles of aluminum oxide formed: Moles of \(Al_2O_3\) formed = (3/2) * Moles of \(Cr_2O_3\) = (3/2) * 0.2237 = 0.3356 mol

Convert moles of aluminum oxide to grams and find the mass of aluminum oxide produced.

Finally, use the molar mass of aluminum oxide to convert the moles of aluminum oxide formed to grams: Mass of \(Al_2O_3\) = (0.3356 mol) * (101.961 g/mol) = 34.2 g So, the mass of aluminum oxide produced in the reaction is approximately 34.2 g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, causing the reaction to stop. It determines the maximum amount of product that can be formed. To identify the limiting reactant, you'll need to compare the mole ratio of the available reactants to the coefficients in the balanced chemical equation.

In our exercise, we compared the moles of chromium(III) oxide and aluminum. By dividing each by the coefficients from the balanced equation, we found that chromium(III) oxide had the lower value when adjusted for its role in the reaction. This meant it was the limiting reactant, dictating the amount of products formed.

• Limiting reactant stops the reaction once used up.
• Comparison of mole ratios helps identify it.

Understanding limiting reactants is key to predicting reaction yields and efficiently utilizing resources.

Molar Mass Calculation

Molar mass is crucial for converting between the mass of a substance and the amount in moles. It’s calculated by summing the atomic masses of all atoms in a molecule, usually found on the periodic table.

For example, in the problem, we calculated the molar mass of several substances:

• For chromium(III) oxide (\(Cr_2O_3\)), we combined the mass of chromium and oxygen: \((2 \times 51.996) + (3 \times 16.000) = 151.992 \, \text{g/mol}\).
• For aluminum (\(Al\)), its molar mass is simply the element’s atomic mass: \(26.981 \, \text{g/mol}\).
• Taking one step further, aluminum oxide (\(Al_2O_3\)) required the sum of the masses of aluminum and oxygen in the compound.

Understanding and calculating molar mass is essential for navigating stoichiometric calculations and further analyzing reactions.

Balanced Chemical Equation

A balanced chemical equation is fundamental in stoichiometry, showing the quantity relationship among reactants and products. It ensures the law of conservation of mass is followed, meaning atoms are neither created nor destroyed during a chemical reaction.

In the exercise's balanced equation:
\[2 \, Cr_2O_3 + 3 \, Al \rightarrow 4 \, Cr + 3 \, Al_2O_3\]

• This balance indicates that two moles of chromium(III) oxide and three moles of aluminum react to produce four moles of chromium and three moles of aluminum oxide.
• Coefficients indicate proportional amounts of each substance.
  To understand a chemical reaction fully, familiarity with its balanced equation is crucial. It serves as the key map for all stoichiometric calculations.

Mole Conversion

Mole conversion is a process that involves changing units from grams to moles (or vice-versa), using the molar mass as a conversion factor. This step is necessary to apply stoichiometry based on the balanced equation.

In our exercise, we converted the masses of Cr₂O₃, Al, and Cr into moles using their molar masses:

• Moles of \(Cr_2O_3\) = \( \frac{34.0}{151.992} = 0.2237 \text{ mol}\)
• Moles of \(Al\) = \( \frac{12.1}{26.981} = 0.4486 \text{ mol}\)
• Moles of \(Cr\) = \( \frac{23.3}{51.996} = 0.4481 \text{ mol}\)

This conversion allows us to utilize mole ratios from the balanced chemical equation effectively. It bridges the gap between measurable quantities in the lab and theoretical calculations necessary for predicting outcomes.