Problem 71 - Chemistry: Principles and Reactions

Question

Consider the hypothetical reaction $$ 8 \mathrm{~A}_{2} \mathrm{~B}_{3}(s)+3 \mathrm{X}_{4}(g) \longrightarrow 4 \mathrm{~A}_{4} \mathrm{X}_{3}(s)+12 \mathrm{~B}_{2}(g) $$ When \(10.0 \mathrm{~g}\) of \(\mathrm{A}_{2} \mathrm{~B}_{3}(\mathrm{MM}=255 \mathrm{~g} / \mathrm{mol})\) react with an excess of \(\mathrm{X}_{4}, 4.00 \mathrm{~g}\) of \(\mathrm{A}_{4} \mathrm{X}_{3}\) are produced. (a) How many moles of \(A_{4} X_{3}\) are produced? (b) What is the molar mass of \(\mathrm{A}_{4} \mathrm{X}_{3}\) ?

Step-by-Step Solution

Answer

Answer: (a) 0.0196 moles of A4X3 are produced, and (b) the molar mass of A4X3 is 204 g/mol.

Step 1: Calculate moles of reactant A2B3

Given the mass and molar mass of A2B3, we can find the moles of A2B3 by using the following formula: Number of moles = (mass of substance) / (molar mass of substance) $$ \text{moles of A}_{2} B_{3} = \frac{10.0 \ g}{255 \ g/mol} = 0.0392 \ mol $$

Step 2: Determine moles of A4X3 using stoichiometry

According to the balanced hypothetical reaction, 8 moles of A2B3 react to produce 4 moles of A4X3. Using the mole ratio, we can determine the moles of A4X3 produced. $$ \text{moles of A}_{4} X_{3} = \text{moles of A}_{2} B_{3} \times \frac{4 \ \text{moles A}_{4} X_{3}}{8 \ \text{moles A}_{2} B_{3}} \\ = 0.0392 \ \text{mol} \times \frac{4}{8} = 0.0196 \ \text{mol} $$ (a) So, 0.0196 moles of \(A_{4}X_{3}\) are produced.

Step 3: Calculate the molar mass of A4X3

Now that we have the moles and the mass of A4X3 produced, we can find its molar mass using the same formula we used in Step 1, rearranged for molar mass: Molar mass of substance = (mass of substance) / (number of moles) $$ \mathrm{MM(A}_{4} \mathrm{X}_{3}) = \frac{4.00 \ g}{0.0196 \ mol} = 204 \ g/mol $$ (b) Finally, the molar mass of \(A_{4}X_{3}\) is \(204 \ g/mol\).

Key Concepts

Molar Mass

Understanding the molar mass of a substance is essential in chemistry, as it relates the mass of a material to the number of particles, typically atoms or molecules, it contains. Expressed in grams per mole (\text{g/mol}), molar mass is derived from the atomic or molecular weight of a substance. It is often calculated by summing the atomic masses of all the atoms in a molecule, as found on the periodic table.

For instance, in the exercise provided, we see the molar mass of compound \textbf{\text{A}}\(_2\) \textbf{\text{B}}\(_3\) is given as 255 g/mol. This information is critical when we want to convert the given mass (in grams) to moles, which is a measure of the number of chemical units of the substance.

Moles Calculation

The concept of moles is at the heart of chemical quantification. One mole represents approximately \(6.022 \times 10^{23}\) entities, be it atoms, ions, or molecules, and is known as Avogadro's number. To calculate the number of moles from mass, the formula is simple:
\[\text{Number of moles} = \frac{\text{mass of substance}}{\text{molar mass of substance}}\]
For the exercise, to find out how many moles of \textbf{\text{A}}\(_4\) \textbf{\text{X}}\(_3\) are produced, we first calculate the moles of reactant \textbf{\text{A}}\(_2\) \textbf{\text{B}}\(_3\) using its given mass and molar mass. This forms the basis to use the stoichiometry of the chemical reaction to find out the moles of product formed, showcasing the importance of accurate mole calculations in predicting the outcomes of reactions.

Chemical Reaction Equation

Chemical reaction equations provide a symbolic representation of the reactants and products involved in a chemical reaction, along with their stoichiometric coefficients, which tell us the proportions in which substances react and are produced. The balanced equation given in the exercise is central to understanding the stoichiometry of the reaction:
\[8 \textbf{A}_2\textbf{B}_3(s) + 3 \textbf{X}_4(g) \longrightarrow 4 \textbf{A}_4\textbf{X}_3(s) + 12 \textbf{B}_2(g)\]
Using stoichiometry, we can relate the quantities of reactants and products through their coefficients. In the illustrated problem, the ratio of the reactant \textbf{\text{A}}\(_2\) \textbf{\text{B}}\(_3\) to the product \textbf{\text{A}}\(_4\) \textbf{\text{X}}\(_3\) is 8:4, simplifying it down to 2:1, meaning for every 2 moles of the reactant, 1 mole of the product is formed. This relationship allows us to calculate the moles of \textbf{\text{A}}\(_4\) \textbf{\text{X}}\(_3\) formed from a given amount of \textbf{\text{A}}\(_2\) \textbf{\text{B}}\(_3\).