Aluminum and chlorine gas react to form aluminum chloride according to the balanced equation shown in below. $$ 2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AlCl}_{3}(s) $$ If \(17.467\) grams of chlorine gas are allowed to react with excess \(\mathrm{Al}\), what mass of solid aluminum chloride will be formed?

Understanding the concept of molar mass is crucial when dealing with chemical reactions and stoichiometry. Molar mass is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It serves as a bridge between the atomic scale and the macroscopic scale. To calculate the molar mass, you add up the atomic masses of all the atoms in the molecule. For example, in the given exercise, chlorine gas (Cl₂) consists of two chlorine atoms. Since the atomic mass of chlorine is 35.45 g/mol, you simply multiply this by two, leading to a molar mass of 70.90 g/mol for Cl₂.

This step is essential and it provides baseline information to move forward with the problem. It allows us to convert from grams to moles, which is a fundamental skill for any subsequent stoichiometric calculation.

Balancing chemical equations is a vital step in solving stoichiometry problems. A balanced equation ensures that the same number of atoms for each element is present on both sides of the reaction, showing the conservation of mass. In our exercise, the chemical equation is already balanced:2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s).

This tells us the exact ratio in which reactants combine and products form, which is crucial for the stoichiometry calculations. For every two moles of aluminum reacted, three moles of chlorine are required, and two moles of aluminum chloride are produced. The coefficients in the equation (2, 3, and 2) directly state the mole ratios, which we then use to predict the amount of product formed from a given amount of reactant. It's important to comprehend this concept thoroughly, as stoichiometry hinges on the relationship between these coefficients.

Once the chemical equation is balanced and the molar mass of the reactants and products determined, we can perform a mole-to-mass conversion. The goal is to convert moles of one substance to grams of another substance using the balanced equation as the conversion factor. From our exercise, after finding the moles of Cl₂, we use the mole ratio of Cl₂ to AlCl₃ from the equation to determine moles of AlCl₃. This is followed by multiplying the moles of AlCl₃ by its calculated molar mass to find the mass in grams of AlCl₃ produced.

Quick Guide

• Identify the number of moles of one substance from the mass given in the problem.
• Use the mole ratio from the balanced equation to find the moles of a product or another reactant.
• Multiply by the molar mass of the product or reactant of interest to find the mass.

Mastering mole-to-mass conversion is crucial, as it allows one to quantitatively predict the outcomes of chemical reactions.