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

In the given reaction, \(34.0 \mathrm{g}\) of chromium(III) oxide reacts with \(12.1 \mathrm{g}\) of aluminum and produces \(23.3 \mathrm{g}\) of chromium. By calculating the number of moles of each substance and using stoichiometry, we can determine that \(34.3 \mathrm{g}\) of aluminum oxide is produced.

Step-by-step solution

Write the balanced chemical equation for the reaction

First, we need to write a balanced chemical equation for the reaction between chromium(III) oxide and aluminum to produce chromium and aluminum oxide: \[2Cr_2O_3 + 3Al \rightarrow 4Cr + 3Al_2O_3\]

Calculate the number of moles of reactants and chromium produced

Now, we need to convert the mass of each reactant and the mass of chromium produced to moles. We will use the molar mass of each compound for this conversion: Molar Masses: - Chromium(III) oxide: \(Cr_2O_3 = 2 * 51.9961 \mathrm{g/mol} (Cr) + 3 * 15.9994 \mathrm{g/mol} (O) = 151.9904 \mathrm{g/mol}\) - Aluminum: \(Al = 26.9815 \mathrm{g/mol}\) - Chromium: \(Cr = 51.9961 \mathrm{g/mol}\) Moles of each substance: - Chromium(III) oxide: \(\frac{34.0 \mathrm{g}}{151.9904 \mathrm{g/mol}} = 0.2236 \mathrm{mol}\) - Aluminum: \(\frac{12.1 \mathrm{g}}{26.9815 \mathrm{g/mol}} = 0.4486 \mathrm{mol}\) - Chromium: \(\frac{23.3 \mathrm{g}}{51.9961 \mathrm{g/mol}} = 0.4481 \mathrm{mol}\)

Use stoichiometry to determine the number of moles of aluminum oxide produced

From the balanced chemical equation (Step 1), we can determine the stoichiometric relationship between the reactants and product: \[2Cr_2O_3 + 3Al \rightarrow 4Cr + 3Al_2O_3\] Using the moles of chromium produced (0.4481 mol), we can determine the number of moles of aluminum oxide produced: \[\frac{3 \mathrm{mol \, Al_2O_3}}{4 \mathrm{mol \, Cr}} * 0.4481 \mathrm{mol \, Cr} = 0.3361 \mathrm{mol \, Al_2O_3}\]

Calculate the mass of aluminum oxide produced

Finally, we can convert the number of moles of aluminum oxide produced to mass using its molar mass: - Aluminum oxide: \(Al_2O_3 = 2 * 26.9815 \mathrm{g/mol} (Al) + 3 * 15.9994 \mathrm{g/mol} (O) = 101.9612 \mathrm{g/mol}\) Mass of aluminum oxide produced: \(0.3361 \mathrm{mol \,Al_2O_3} * 101.9612 \mathrm{g/mol} = 34.3 \mathrm{g}\) So, 34.3 g of aluminum oxide is produced.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is crucial for understanding chemical reactions. It represents the reactants and products in a straightforward way. Each element present in the reactants must also be in the products with equal numbers of atoms. This balance is necessary due to the law of conservation of mass, which states that mass is conserved in any given chemical reaction. So in the equation\[2Cr_2O_3 + 3Al \rightarrow 4Cr + 3Al_2O_3\],the coefficients reflect how many moles of each reactant participate to yield the products. Each side of the equation must have the same number of atoms for every element involved. For instance, there are six chromium and six oxygen atoms on each side of this reaction. Balancing chemical equations ensures that we accurately describe what occurs during a reaction and serves as the basis for more complex stoichiometric calculations.

Mole Concept

The mole is a fundamental concept in chemistry that bridges the gap between the microscopic world of atoms and the macroscopic world we observe. It allows chemists to count atoms and molecules by weighing them. One mole is 6.022 \times 10^{23} units of a substance and is called Avogadro's number. This makes it similar to a dozen but much larger.

In our reaction example, calculating the moles of chromium(III) oxide and aluminum tells us how many molecules of each substance we have. This is done by dividing the given mass by the molar mass of the substance. For instance:

• Moles of Chromium(III) oxide: \( \frac{34.0 \text{ g}}{151.9904 \text{ g/mol}} = 0.2236 \text{ mol} \)

The mole allows chemists to use balanced equations to predict the outcomes of reactions quantitatively, which means we can determine the amount of products formed or the reactants required to achieve a reaction.

Molar Mass Calculation

Molar mass is the weight of one mole of a substance. It provides an essential conversion factor between the mass of a substance and the quantity of substance. The molar mass is crucial for converting grams to moles, enabling us to use stoichiometry accurately.

To perform molar mass calculations, you need the atomic masses of the elements from the periodic table. For example, the molar mass of chromium(III) oxide is calculated by adding the masses of all atoms in the formula:

• \(Cr_2O_3 = 2 \times 51.9961 \text{ g/mol (Cr)} + 3 \times 15.9994 \text{ g/mol (O)} = 151.9904 \text{ g/mol}\)

By knowing the molar mass, you can convert the mass of a substance in grams into moles, which aids in stoichiometric calculations. This step is essential when determining how much of a product forms or how much reactant is necessary, as seen in calculating the mass of aluminum oxide produced from a given reaction.