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Chapter 20: Problem 27

Boron hydrides were once evaluated for possible use as rocket fuels. Complete and balance the following equation for the combustion of diborane. $$ \mathrm{B}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{B}(\mathrm{OH})_{3}(s) $$

Short Answer

Expert verified
The balanced equation for the combustion of diborane with oxygen gas is: \( \mathrm{B}_{2} \mathrm{H}_{6}(g)+3\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{B}(\mathrm{OH})_{3}(s) \).

Step by step solution

01

Write the unbalanced equation

Start with the unbalanced equation: \( \mathrm{B}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{B}(\mathrm{OH})_{3}(s) \)
02

Inspect the number of atoms of each element

Find the number of atoms of each element on both sides of the equation: - On the reactant side: 2 boron (B), 6 hydrogen (H), and 2 oxygen (O) - On the product side: 1 boron (B), 3 hydrogen (H), and 3 oxygen (O)
03

Balance the boron atoms

To balance the boron atoms, we need two B(OH)₃ molecules on the product side: \( \mathrm{B}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{B}(\mathrm{OH})_{3}(s) \) Now we have: - On the reactant side: 2 boron (B), 6 hydrogen (H), and 2 oxygen (O) - On the product side: 2 boron (B), 6 hydrogen (H), and 6 oxygen (O)
04

Balance the hydrogen atoms

The hydrogen atoms are already balanced, so no adjustments are needed.
05

Balance the oxygen atoms

To balance the oxygen atoms, we need to have 6 oxygen atoms provided by O₂ on the reactant side: \( \mathrm{B}_{2} \mathrm{H}_{6}(g)+3\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{B}(\mathrm{OH})_{3}(s) \) Now the equation is balanced: - On the reactant side: 2 boron (B), 6 hydrogen (H), and 6 oxygen (O) - On the product side: 2 boron (B), 6 hydrogen (H), and 6 oxygen (O)

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Most popular questions from this chapter

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