Question

Write balanced equations for the following reactions: (a) boron trichloride with water, (b) cobalt (II) oxide with nitric acid, (c) phosphorus pentoxide with water, (d) carbon dioxide with aqueous barium hydroxide.

Short answer

The balanced equations for the given reactions are: (a) BCl3 + 3 H2O → H3BO3 + 3 HCl (b) CoO + 2 HNO3 → Co(NO3)2 + H2O (c) P4O10 + 6 H2O → 4 H3PO4 (d) CO2 + Ba(OH)2 → BaCO3 + H2O

Step-by-step solution

(Reaction a: Boron trichloride with water)

In this reaction, boron trichloride (BCl3) reacts with water (H2O). This is an example of a decomposition reaction where boron trichloride hydrolyzes in the presence of water, forming boric acid (H3BO3) and hydrochloric acid (HCl). Step 1. Write the unbalanced equation: BCl3 + H2O → H3BO3 + HCl Step 2. Balancing the equation: To balance the equation, adjust the coefficients such that the number of atoms on the reactant side equals the number of atoms on the product side. BCl3 + 3 H2O → H3BO3 + 3 HCl Hence, the balanced equation for the reaction between boron trichloride and water is BCl3 + 3 H2O → H3BO3 + 3 HCl.

(Reaction b: Cobalt (II) oxide with nitric acid)

In this reaction, cobalt (II) oxide (CoO) reacts with nitric acid (HNO3). This is a double displacement reaction involving an oxide and an acid, resulting in the formation of a salt, cobalt (II) nitrate (Co(NO3)2), and water (H2O). Step 1. Write the unbalanced equation: CoO + HNO3 → Co(NO3)2 + H2O Step 2. Balancing the equation: To balance the equation, adjust the coefficients such that the number of atoms on the reactant side equals the number of atoms on the product side. CoO + 2 HNO3 → Co(NO3)2 + H2O Hence, the balanced equation for the reaction between cobalt (II) oxide and nitric acid is CoO + 2 HNO3 → Co(NO3)2 + H2O.

(Reaction c: Phosphorus pentoxide with water)

In this reaction, phosphorus pentoxide (P4O10) reacts with water (H2O). This is a synthesis reaction resulting in the formation of phosphoric acid (H3PO4). Step 1. Write the unbalanced equation: P4O10 + H2O → H3PO4 Step 2. Balancing the equation: To balance the equation, adjust the coefficients such that the number of atoms on the reactant side equals the number of atoms on the product side. P4O10 + 6 H2O → 4 H3PO4 Hence, the balanced equation for the reaction between phosphorus pentoxide and water is P4O10 + 6 H2O → 4 H3PO4.

(Reaction d: Carbon dioxide with aqueous barium hydroxide)

In this reaction, carbon dioxide (CO2) reacts with aqueous barium hydroxide (Ba(OH)2). This is another example of a double displacement reaction, resulting in the formation of barium carbonate (BaCO3) and water (H2O). Step 1. Write the unbalanced equation: CO2 + Ba(OH)2 → BaCO3 + H2O Step 2. Balancing the equation: To balance the equation, adjust the coefficients such that the number of atoms on the reactant side equals the number of atoms on the product side. CO2 + Ba(OH)2 → BaCO3 + H2O Here, the equation is already balanced with equal numbers of atoms on both reactant and product sides, so no change is needed in coefficients. Hence, the balanced equation for the reaction between carbon dioxide and aqueous barium hydroxide is CO2 + Ba(OH)2 → BaCO3 + H2O.

Key concepts

Balancing Equations

Balancing chemical equations is crucial. It ensures the conservation of mass, as atoms are neither created nor destroyed in chemical reactions. To balance an equation, adjust the coefficients (numbers before the chemical formulas) to have equal numbers of each type of atom on both sides of the equation. For example, in the reaction between boron trichloride and water, the unbalanced equation is: \( \text{BCl}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + \text{HCl} \). By adjusting coefficients to preserve the number of chlorine, oxygen, and boron atoms, the balanced equation becomes: \( \text{BCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + 3 \text{HCl} \). This balancing principle applies to all types of chemical reactions, ensuring the fundamental law of conservation is respected.

Decomposition Reactions

Decomposition reactions involve a single compound breaking down into two or more simpler substances. In the context of the provided exercise, they occur when boron trichloride reacts with water. The reaction can be summarized as: \( \text{BCl}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + \text{HCl} \), which features boron trichloride (a single reactant) breaking down to form boric acid and hydrochloric acid. These reactions can be prompted by heat, light, or other chemical catalysts, and they are often endothermic, meaning they absorb energy. Understanding decomposition reactions is pivotal in fields ranging from industrial chemistry to environmental science.

Synthesis Reactions

Synthesis reactions involve combining two or more substances to form a single, more complex compound. The reaction between phosphorus pentoxide and water in the exercise is an example: \( \text{P}_4\text{O}_{10} + 6 \text{H}_2\text{O} \rightarrow 4 \text{H}_3\text{PO}_4 \). Here, phosphorus pentoxide and water combine to form phosphoric acid. Synthesis reactions are essential for creating compounds and materials in laboratories and industries. They often require energy input and can be exothermic (release energy) or endothermic. Familiarity with these reactions is essential for chemists in creating new materials and products.

Double Displacement Reactions

Double displacement reactions typically involve two compounds exchanging ions to form two new compounds. For instance, in the reaction between cobalt (II) oxide and nitric acid, we have: \( \text{CoO} + 2 \text{HNO}_3 \rightarrow \text{Co(NO}_3\text{)}_2 + \text{H}_2\text{O} \). Or, when carbon dioxide reacts with barium hydroxide, resulting in \( \text{CO}_2 + \text{Ba(OH)}_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} \). These reactions are common in chemistry for neutralization, precipitation, and complexation processes. Understanding them is crucial for predicting reaction outcomes and manipulating chemical pathways.