Question

For each of the following precipitation reactions, complete and balance the equation, indicating clearly which product is the precipitate. a. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(a q)+\mathrm{CoCl}_{2}(a q) \rightarrow\) b. \(\operatorname{FeCl}_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow\) c. \(\operatorname{CuSO}_{4}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow\)

Short answer

a. \( (\mathrm{NH}_{4})_{2} \mathrm{S}(a q) + \mathrm{CoCl}_{2}(a q) \rightarrow 2\mathrm{NH}_{4}\mathrm{Cl}(a q) + \mathrm{CoS}(s) \) b. \( \operatorname{FeCl}_{3}(a q) + 3\mathrm{NaOH}(a q) \rightarrow \mathrm{Fe(OH)}_{3}(s) + 3\mathrm{NaCl}(a q) \) c. \( \operatorname{CuSO}_{4}(a q) + \mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{CuCO}_{3}(s) + \mathrm{Na}_{2}\mathrm{SO}_{4}(a q) \)

Step-by-step solution

Identify the double displacement reactions

Double displacement reactions involve the exchange of ions between two compounds. In each case, two aqueous solutions are mixed, and a solid precipitate forms. The general form of a double displacement reaction is: $$ \mathrm{AB}(a q)+\mathrm{CD}(a q) \rightarrow \mathrm{AC}(a q, s)+\mathrm{BD}(a q, s) $$

Determine the precipitate using solubility rules

To determine which product is the precipitate, we will use some common solubility rules: 1. Most alkali metal and ammonium salts are soluble. 2. Nitrates, acetates, and chlorates are soluble. 3. Most chloride, bromide, and iodide salts are soluble (exceptions: with Ag+, Hg2+, and Pb2+ ions). 4. Sulfates are soluble (exceptions: with Ba2+, Ca2+, Sr2+, Hg2+, and Pb2+ ions). 5. Carbonates, phosphates, and sulfides are generally insoluble (exceptions: with alkali metals and ammonium ions).

Reaction a: \((\mathrm{NH}_{4})_{2} \mathrm{S}(a q)+\mathrm{CoCl}_{2}(a q) \rightarrow\)

Following the steps mentioned above, the double displacement reaction will give the products as follows: $$ (\mathrm{NH}_{4})_{2} \mathrm{S}(a q) + \mathrm{CoCl}_{2}(a q) \rightarrow \mathrm{NH}_{4}\mathrm{Cl}(a q) + \mathrm{CoS}(s) $$ The product \(\mathrm{CoS}\) is insoluble, so it will be the precipitate. Now, balance the equation: $$ (\mathrm{NH}_{4})_{2} \mathrm{S}(a q) + \mathrm{CoCl}_{2}(a q) \rightarrow 2\mathrm{NH}_{4}\mathrm{Cl}(a q) + \mathrm{CoS}(s) $$

Reaction b: \(\operatorname{FeCl}_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

Following the steps mentioned above, the double displacement reaction will give the products as follows: $$ \operatorname{FeCl}_{3}(a q) + \mathrm{NaOH}(a q) \rightarrow \mathrm{Fe(OH)}_{3}(s) + \mathrm{NaCl}(a q) $$ The product \(\mathrm{Fe(OH)}_{3}\) is insoluble, so it will be the precipitate. Now, balance the equation: $$ \operatorname{FeCl}_{3}(a q) + 3\mathrm{NaOH}(a q) \rightarrow \mathrm{Fe(OH)}_{3}(s) + 3\mathrm{NaCl}(a q) $$

Reaction c: $\operatorname{CuSO}_{4}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow$

Following the steps mentioned above, the double displacement reaction will give the products as follows: $$ \operatorname{CuSO}_{4}(a q) + \mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{CuCO}_{3}(s) + \mathrm{Na}_{2}\mathrm{SO}_{4}(a q) $$ The product \(\mathrm{CuCO}_{3}\) is insoluble, so it will be the precipitate. Now, balance the equation: $$ \operatorname{CuSO}_{4}(a q) + \mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{CuCO}_{3}(s) + \mathrm{Na}_{2}\mathrm{SO}_{4}(a q) $$

Key concepts

Double Displacement Reactions

Double displacement reactions, also known as metathesis reactions, occur when two compounds exchange ions to form two new compounds. These reactions typically involve aqueous solutions that result in the formation of a solid precipitate. Imagine mixing two clear solutions, and suddenly, a cloud of particles forms. That's the magic of double displacement reactions!

In a double displacement reaction, the general formula is:

• \( \text{AB}(aq) + \text{CD}(aq) \rightarrow \text{AD}(aq, s) + \text{CB}(aq, s) \)

Here, compounds \(\text{AB}\) and \(\text{CD}\) swap their partners to create \(\text{AD}\) and \(\text{CB}\), where one or both may form a precipitate.

This type of reaction is especially important in predicting product formation in chemistry labs and industrial processes.

Solubility Rules

Solubility rules are guidelines used to predict whether a compound will dissolve in water. When dealing with precipitation reactions, these rules help you determine which of the resulting products will be insoluble and form a precipitate.

Here's a quick guide to some common solubility rules:

• Alkali metal and ammonium (\(\text{NH}_4^+\)) salts are generally soluble.
• Nitrate, acetate, and chlorate salts dissolve readily.
• Chlorides, bromides, and iodides are soluble, except when paired with ions like \(\text{Ag}^+\), \(\text{Hg}_2^{2+}\), and \(\text{Pb}^{2+}\).
• Sulfates are soluble, with exceptions like \(\text{Ba}^{2+}\), \(\text{Ca}^{2+}\), \(\text{Sr}^{2+}\), \(\text{Hg}_2^{2+}\), and \(\text{Pb}^{2+}\).
• Carbonates, phosphates, and sulfides are typically insoluble, except with alkali metals and ammonium.

By applying these rules, we can identify the precipitate in a reaction, helping us to better understand the reaction's outcomes.

Chemical Equation Balancing

Balancing chemical equations is essential to ensure that the same number of each type of atom appears on both sides of a reaction. This reflects the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction.

Steps to balance a chemical equation:

• Write the unbalanced equation.
• Count the atoms of each element on both sides of the equation.
• Use coefficients to balance each element, starting with those that appear in only one reactant and product.
• Double-check the balance of atoms to ensure accuracy.

For example, in the reaction of \[ (\text{NH}_4)_2\text{S}(aq) + \text{CoCl}_2(aq) \rightarrow 2\text{NH}_4\text{Cl}(aq) + \text{CoS}(s) \]The balanced equation accurately reflects the number of each atom involved, ensuring the equation is consistent with chemical laws.

Balancing equips you with the ability to predict reactant amounts and yields of products in practical chemical processes.