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Chapter 22: Problem 78

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.

Short Answer

Expert verified
\( (a) \: B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 3H_2 \) \( (b) \: 4H_3BO_3 \rightarrow H_2B_4O_7 + 3H_2O \) \( (c) \: B_2O_3 + 3H_2O \rightarrow 2H_3BO_3 \)

Step by step solution

01

Reaction (a): Diborane reacts with water

We need to write a balanced equation for diborane (B2H6) reacting with water (H2O) to form boric acid (H3BO3) and molecular hydrogen (H2). Unbalanced equation: B2H6 + H2O -> H3BO3 + H2 Now, balance the equation by adjusting the coefficients: B2H6 + 6H2O -> 2H3BO3 + 3H2 Balanced equation: B2H6 + 6H2O -> 2H3BO3 + 3H2
02

Reaction (b): Boric acid undergoes a condensation reaction

We need to write a balanced equation for boric acid (H3BO3) undergoing a condensation reaction to form tetraboric acid (H2B4O7). Unbalanced equation: H3BO3 -> H2B4O7 For a condensation reaction, multiple units of boric acid will combine to form tetraboric acid. In this case, we need 4 units of boric acid to form 1 unit of tetraboric acid and 3 molecules of water, as they are released during the reaction. Balanced equation: 4H3BO3 -> H2B4O7 + 3H2O Reaction (c): Boron oxide dissolves in water to give a solution of boric acid.
03

Reaction (c): Boron oxide dissolves in water

We need to write a balanced equation for boron oxide (B2O3) dissolving in water (H2O) to form a solution of boric acid (H3BO3). Unbalanced equation: B2O3 + H2O -> H3BO3 Now, balance the equation by adjusting the coefficients: B2O3 + 3H2O -> 2H3BO3 Balanced equation: B2O3 + 3H2O -> 2H3BO3

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Most popular questions from this chapter

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