Question

Elemental bromine is the source of bromine compounds. The element is produced from certain brine solutions that occur naturally. These brines are essentially solutions of calcium bromide that, when treated with chlorine gas, yield bromine in a displacement reaction. What are the molecular equation and net ionic equation for the reaction? A solution containing \(40.0 \mathrm{~g}\) of calcium bromide requires \(14.2 \mathrm{~g}\) of chlorine to react completely with it, and \(22.2 \mathrm{~g}\) of calcium chloride is produced in addition to whatever bromine is obtained. How many grams of calcium bromide are required to produce 10.0 pounds of bromine?

Short answer

5715.36 grams of \(\text{CaBr}_2\) are required to produce 10.0 pounds of bromine.

Step-by-step solution

Write the Molecular Equation

The molecular equation for the reaction between calcium bromide \( \text{CaBr}_2 \) and chlorine gas \( \text{Cl}_2 \) can be written as follows:\[ \text{CaBr}_2 + \text{Cl}_2 \rightarrow \text{CaCl}_2 + \text{Br}_2 \]

Write the Net Ionic Equation

To write the net ionic equation, we split all the strong electrolytes (aqueous compounds) into their ions. Calcium bromide is soluble and chlorine gas and bromine are diatomic molecules. The net ionic equation is:\[ \text{Br}^- + \text{Cl}_2 \rightarrow \text{Cl}^- + \text{Br}_2 \]Here, calcium ions are spectator ions and do not appear in the net ionic equation.

Convert Pounds to Grams

Since we want to produce 10.0 pounds of bromine (\( \text{Br}_2 \)), we first convert pounds to grams. 1 pound = 453.592 grams, therefore 10.0 pounds = 10.0 \times 453.592 = 4535.92 grams of bromine.

Use Stoichiometry to Find Mass of \(\text{CaBr}_2\)

Using the molecular equation, we see that 1 mole of \(\text{CaBr}_2\) produces 1 mole of \(\text{Br}_2\).- Molar mass of \(\text{CaBr}_2 = 40.08 + 2 \times 79.904 = 199.88 \text{ g/mol}\)- Molar mass of \(\text{Br}_2 = 2 \times 79.904 = 159.808 \text{ g/mol}\)Using stoichiometry:\[ \text{g CaBr}_2 = 199.88 \times \left( \frac{4535.92}{159.808} \right) = 5715.36 \text{ grams} \]

Key concepts

Calcium Bromide

Calcium bromide, represented as \(\text{CaBr}_2\), is a chemical compound composed of calcium and bromine. It appears as a colorless crystalline solid at room temperature and is highly soluble in water.
In reactions involving calcium bromide, the compound often dissociates completely in solution because it is a strong electrolyte. This property is important because it enables calcium bromide to participate effectively in various chemical reactions, including displacement reactions.

• Formula: \(\text{CaBr}_2\)
• Molar Mass: \(40.08 + 2 \times 79.904 = 199.88\) g/mol

When calcium bromide is treated with chlorine gas, as described in the original exercise, a displacement reaction occurs. This reaction highlights the reactivity of chlorine gas, which displaces bromine from calcium bromide, forming calcium chloride and elemental bromine. This is why calcium bromide is considered a crucial starting material in the production of bromine.

Net Ionic Equation

Net ionic equations focus on the species that actually participate in the reaction, excluding spectator ions. In the reaction between calcium bromide \((\text{CaBr}_2)\) and chlorine gas \((\text{Cl}_2)\), determining the net ionic equation involves identifying the ions that change during the reaction.
Firstly, calcium bromide dissolves into its ions: calcium ions \((\text{Ca}^{2+})\) and bromide ions \((\text{Br}^-)\). Chlorine gas is a diatomic molecule and does not dissociate into ions in the same way. This results in the net ionic equation:

• \(\text{Br}^- + \text{Cl}_2 \rightarrow \text{Cl}^- + \text{Br}_2\)

In this equation, bromine ions \((\text{Br}^-)\) react with chlorine gas \((\text{Cl}_2)\) to form chloride ions \((\text{Cl}^-)\) and bromine \((\text{Br}_2)\). Calcium ions \((\text{Ca}^{2+})\) remain unchanged and are therefore omitted from the net ionic equation, highlighting their role as spectator ions. This method provides a clearer view of the essential chemical changes occurring during the reaction.

Stoichiometry

Stoichiometry is the field of chemistry that focuses on the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced in a reaction.
The original problem uses stoichiometry to determine how many grams of calcium bromide are needed to produce a given quantity of bromine. First, the reaction's balanced molecular equation is used to determine the molar relationship between the substances. By comparing the molar mass of calcium bromide \((199.88 \text{ g/mol})\) and bromine \((159.808 \text{ g/mol})\), we can calculate the mass of calcium bromide required to produce 10.0 pounds of bromine.

• Conversion from pounds to grams is essential: 10.0 pounds \(= 4535.92\) grams.
• The molar ratios are based on a one-to-one relationship between \(\text{CaBr}_2\) and \(\text{Br}_2\).

With stoichiometry, we find that 5715.36 grams of calcium bromide are required. This thorough understanding of reactant-product relationships makes stoichiometry a powerful tool in chemical calculations.