Problem 17 - Chemistry The Molecular Science

Question

Balance this equation and determine which box represents reactants and which box represents products. $$\mathrm{Sb}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SbCl}_{3}(\mathrm{~g})$$

Step-by-Step Solution

Answer

Reactants: Sb and Cl₂; Product: SbCl₃; Balanced equation: 2Sb + 3Cl₂ → 2SbCl₃.

Step 1: Identify Each Element

Make a list of all elements found in the chemical equation: Sb (Antimony) and Cl (Chlorine).

Step 2: Count Atoms on Each Side

Count the number of atoms for each element on both sides of the equation. On the left, we have 1 Sb and 2 Cl atoms, and on the right, in SbCl₃, we have 1 Sb and 3 Cl atoms.

Step 3: Balance the Chlorine Atoms

To balance the Cl atoms, place a coefficient of 1.5 (or use fractions if needed for intermediate steps) before Cl₂ on the reactants side so that there are 3 Cl atoms. This gives us: \[\mathrm{Sb}(\mathrm{g})+1.5\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SbCl}_{3}(\mathrm{~g})\].

Step 4: Adjust to Whole Numbers

Multiplying the entire equation by 2 gets rid of any fractions: \[2\mathrm{Sb}(\mathrm{g})+3\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2\mathrm{SbCl}_{3}(\mathrm{~g})\]. Now we have whole numbers and there are 2 Sb and 6 Cl atoms on each side.

Step 5: Verify Atom Balance

Double-check that both Sb and Cl are balanced. On the left, there are 2 Sb and 6 Cl, and on the right, the same number of both types of atoms are present, satisfying the conservation of mass.

Step 6: Identify Reactants and Products

The substances on the left side of the chemical equation (Sb and Cl₂ gases) are reactants, and the substance on the right side (SbCl₃ gas) is the product.

Key Concepts

Reactants and Products

In the world of chemistry, chemical reactions involve substances transforming into others. These substances onboard each side of the equation are termed reactants and products. The reactants, present on the left side of the equation, are substances that undergo change. For our example, the reactants are Antimony (\( \text{Sb} \)) and Chlorine (\( \text{Cl}_2 \)) gases. Products, found on the equation's right side, are the end findings of the reaction.
In this exercise, Antimony Trichloride (\( \text{SbCl}_3 \)) gas is the resulting product. This output occurs as the reactants recombine their atoms, forming new bonds and thus, new substances that display different properties compared to the starting elements. Understanding which substances are reactants and products is key to grasping how chemical reactions transform matter.

Conservation of Mass

The principle of the conservation of mass tells us that mass cannot be created or destroyed in a chemical reaction. This fundamental notion implies that whatever mass we start with in the reactants will equal the mass found in the products.
When we balance chemical equations, like the one provided, we ensure equal numbers of atoms exist on both sides, confirming the mass balance. In our exercise, we started with 1 Antimony and 2 Chlorine atoms on the reactant side. To get the equally balanced equation, we adjusted it to 2 Antimony and 6 Chlorine atoms on each side, aligning with conservation principles.
Balancing ensures the atom count remains consistent. By using coefficients to adjust the amounts of each substance, we effectively maintain a stable mass flow across the reaction.

Stoichiometry

Stoichiometry, a core concept in chemistry, deals with the quantitative relationships within a chemical reaction. It relies heavily on mole ratios derived from balanced equations to determine the relationships between reactants and products.
In our given example, balancing the equation results in the stoichiometric ratio 2:3:2 for \( \text{Sb} \): \( \text{Cl}_2 \): \( \text{SbCl}_3 \). This ratio tells us how many units of one material will react with a specific number of units of another to form a product.
The calculated whole number coefficients vividly convey this relationship, ensuring that if we start with 2 Antimony and 3 Chlorine molecules, we will end up forming 2 molecules of Antimony Trichloride. This predictive capability is at the heart of stoichiometry, enabling chemists to calculate required and expected amounts efficiently.