Problem 82 - Chemistry: The Central Science

Question

Write a balanced equation for each of the following reactions: (a) Diborane reacts with water to form boric acid and molecular hydrogen. (b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid. (c) Boron oxide dissolves in water to give a solution of boric acid.

Step-by-Step Solution

Answer

(a) \(B_2H_6 (g) + 6H_2O (l) \rightarrow 2H_3BO_3 (aq) + 3H_2 (g)\) (b) \(4H_3BO_3 (s) \rightarrow H_2B_4O_7 (s) + 5H_2O (g)\) (c) \(B_2O_3 (s) + 3H_2O (l) \rightarrow 2H_3BO_3 (aq)\)

Step 1: Reaction (a): Diborane reacts with water to form boric acid and molecular hydrogen.

First, we find the chemical formulas of the reactants and products: Diborane (B2H6), water (H2O), boric acid (H3BO3), and molecular hydrogen (H2). Now, we balance the equation: B2H6 (g) + 6H2O (l) → 2H3BO3 (aq) + 3H2 (g)

Step 2: Reaction (b): Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid.

First, we find the chemical formulas of the reactants and products: Boric acid (H3BO3) and tetraboric acid (H2B4O7). Now, we balance the equation: 4H3BO3 (s) → H2B4O7 (s) + 5H2O (g)

Step 3: Reaction (c): Boron oxide dissolves in water to give a solution of boric acid.

First, we find the chemical formulas of the reactants and products: Boron oxide (B2O3) and boric acid (H3BO3). Now, we balance the equation: B2O3 (s) + 3H2O (l) → 2H3BO3 (aq)