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Chapter 20: Problem 33

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 is: \( B_2H_6(g) + 3O_2(g) \longrightarrow 2BOH_3(s) \)

Step by step solution

01

Write the unbalanced equation

Start by writing down the unbalanced equation as given: \( B_2H_6(g) + O_2(g) \longrightarrow BOH_3(s) \)
02

Balance elements other than O and H

The first step in balancing the equation is to balance elements other than Oxygen and Hydrogen. In this case, there is only 1 element other than Oxygen and Hydrogen, and that is Boron. Notice that there are 2 Boron atoms on the left-hand side (in the B₂H₆ molecule) and only 1 Boron atom on the right-hand side (in the BOH₃ molecule). To balance the number of Boron atoms, add a coefficient of 2 in front of the BOH₃ molecule: \( B_2H_6(g) + O_2(g) \longrightarrow 2BOH_3(s) \)
03

Balance hydrogen atoms

Next, let's balance the Hydrogen atoms. There are 6 Hydrogen atoms on the left-hand side (in the B₂H₆ molecule) and 6 Hydrogen atoms on the right-hand side (in the two BOH₃ molecules). Since the number of Hydrogen atoms is already balanced on both sides, we don't need to make any adjustments here.
04

Balance oxygen atoms

Lastly, let's balance the Oxygen atoms. There are no Oxygen atoms on the left-hand side and 6 Oxygen atoms on the right-hand side (in the two BOH₃ molecules). To balance the number of Oxygen atoms, add a coefficient of 3 in front of the O₂ molecule: \( B_2H_6(g) + 3O_2(g) \longrightarrow 2BOH_3(s) \)
05

Check the balanced equation

Verify that the given equation is properly balanced by confirming that there are the same number of atoms for each element on both sides of the equation. There are 2 Boron atoms, 6 Hydrogen atoms, and 6 Oxygen atoms on both sides of the balanced equation, so the equation is balanced. The final balanced equation is: \( B_2H_6(g) + 3O_2(g) \longrightarrow 2BOH_3(s) \)

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Most popular questions from this chapter

The Group 5 \(\mathrm{A}\) elements can form molecules or ions that involve three, five, or six covalent bonds; \(\mathrm{NH}_{3}, \mathrm{AsCl}_{5},\) and \(\mathrm{PF}_{6}-\) are examples. Draw the Lewis structure for each of these substances, and predict the molecular structure and hybridization for each. Why doesn't \(\mathrm{NF}_{5}\) or \(\mathrm{NCl}_{6}-\) form?

When sodium reacts with hydrogen gas, sodium hydride is produced. Is sodium hydride an ionic or a covalent compound? When sodium hydride reacts with water, the equation is: $$ \mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)+\mathrm{H}_{2}(g) $$ Show that this reaction can be considered both an oxidationreduction reaction and an acid-base reaction.

A cylinder fitted with a movable piston initially contains 2.00 moles of \(\mathrm{O}_{2}(g)\) and an unknown amount of \(\mathrm{SO}_{2}(g) .\) The oxygen is known to be in excess. The density of the mixture is 0.8000 $\mathrm{g} / \mathrm{L}\( at some \)T\( and \)P$ . After the reaction has gone to completion, forming \(\mathrm{SO}_{3}(g),\) the density of the resulting gaseous mixture is 0.8471 \(\mathrm{g} / \mathrm{L}\) at the same \(T\) and \(P\) . Calculate the mass of \(\mathrm{SO}_{3}\) formed in the reaction.

EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt $\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}$ , are also used to treat heavy metal poisoning. The equilibrium constant for the following reaction is \(6.7 \times 10^{21} :\) Calculate \(\left[\mathrm{Pb}^{2+}\right]\) at equilibrium in a solution originally 0.0050 \(\mathrm{M}\) in \(\mathrm{Pb}^{2+}, 0.075 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-},\) and buffered at \(\mathrm{pH}=7.00 .\)

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