Question

Diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) is a gas at room temperature that forms explosive mixtures with air. It reacts with oxygen according to the following equation: \(\mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)\) How many grams of diborane will react with \(7.5 \mathrm{~mol}\) of \(\mathrm{O}_{2} ?\)

Short answer

69.1 grams of diborane will react with 7.5 moles of \(\mathrm{O}_{2}\).

Step-by-step solution

Understand the Stoichiometry

The balanced chemical equation is: \[ \mathrm{B}_{2} \mathrm{H}_{6}(g) + 3 \mathrm{O}_{2}(g) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s) + 3 \mathrm{H}_{2} \mathrm{O}(l) \]This equation tells us that 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\) reacts with 3 moles of \(\mathrm{O}_{2}\).

Set Up Mole Ratio

The mole ratio from the equation is 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\) for every 3 moles of \(\mathrm{O}_{2}\). Given \(7.5\) moles of \(\mathrm{O}_{2}\), use the ratio: \[ \frac{1\, \text{mol} \; \mathrm{B}_{2} \mathrm{H}_{6}}{3\, \text{mol} \; \mathrm{O}_{2}} \] Thus, moles of \(\mathrm{B}_{2} \mathrm{H}_{6}\) needed are \( \frac{7.5}{3} = 2.5 \) moles.

Calculate Molar Mass

The molar mass of \(\mathrm{B}_{2} \mathrm{H}_{6}\) is calculated as follows:- Boron (B): \(10.81\, \text{g/mol}\) - Hydrogen (H): \(1.01\, \text{g/mol}\)Thus,\[ 2(10.81) + 6(1.01) = 27.64 \, \text{g/mol} \]

Calculate Mass of Diborane

Using the moles of \(\mathrm{B}_{2} \mathrm{H}_{6}\) and its molar mass, calculate the mass:\[ \text{Mass} = 2.5 \, \text{mol} \times 27.64 \, \text{g/mol} = 69.1 \, \text{g} \]

Key concepts

Chemical Reactions

In chemistry, a chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Understanding a chemical reaction involves knowing the reactants, which are the substances that enter the reaction, and the products, which are the substances formed as a result of the reaction. Each reaction can be described using a chemical equation.

In the case of diborane (\(\mathrm{B}_{2} \mathrm{H}_{6}\)) reacting with oxygen (\(\mathrm{O}_{2}\)), the balanced chemical equation is: \[\mathrm{B}_{2} \mathrm{H}_{6}(g) + 3 \mathrm{O}_{2}(g) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s) + 3 \mathrm{H}_{2} \mathrm{O}(l) \]

This tells us that one mole of diborane reacts with three moles of oxygen to produce one mole of boron oxide (\(\mathrm{B}_{2} \mathrm{O}_{3}\)) and three moles of water (\(\mathrm{H}_{2} \mathrm{O}\)). It's crucial to balance chemical equations, as it reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

Mole Ratio

The concept of a mole ratio is fundamental in stoichiometry, which is the part of chemistry that deals with the quantitative relationships of substances as they participate in chemical reactions. A mole ratio is derived from the coefficients of a balanced chemical equation. It tells you how many moles of each substance are involved in the reaction.

In our diborane and oxygen reaction, the balanced equation indicates a mole ratio of 1:3 between diborane and oxygen. This means one mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\) reacts with three moles of \(\mathrm{O}_{2}\). Given 7.5 moles of oxygen, we use this ratio to determine that 2.5 moles of diborane will be required to react with it completely. The calculation is straightforward: \[\frac{7.5}{3} = 2.5\]

Understanding mole ratios will enable you to predict how much of a reactant is needed or how much product will be formed in a reaction.

Molar Mass Calculation

Calculating the molar mass of a compound is essential when solving stoichiometry problems, as it allows you to convert between moles and grams. Molar mass is the mass of one mole of a given substance (chemical element or chemical compound) and is expressed in units of grams per mole (g/mol).

For diborane (\(\mathrm{B}_{2} \mathrm{H}_{6}\)), the molar mass is calculated by summing the molar masses of its constituting elements. Boron has a molar mass of approximately 10.81 g/mol, and hydrogen has a molar mass of approximately 1.01 g/mol. Thus, the calculation for diborane's molar mass is:

• 2 Boron atoms: \(2 \times 10.81 = 21.62 \, \text{g/mol}\)
• 6 Hydrogen atoms: \(6 \times 1.01 = 6.06 \, \text{g/mol}\)

Adding these gives the molar mass of diborane:
\[\text{Molar Mass} = 21.62 + 6.06 = 27.68 \, \text{g/mol}\]

Diborane Chemistry

Diborane (\(\mathrm{B}_{2} \mathrm{H}_{6}\)) is a fascinating compound in the field of chemistry known for its unique bonding and reactivity. This compound consists of boron and hydrogen atoms and is considered an electron-deficient compound, which means it does not have enough electrons to conventionally form stable bonds between each atom.

Diborane is mostly studied under organometallic chemistry and is known for its ability to form bridges between atoms, creating a distinct and often studied "banana bond." This contributes to the unique three-center two-electron bonds involving two boron atoms and a hydrogen atom bridged between them. Apart from its intriguing structure, diborane is highly reactive and forms explosive mixtures with air, as seen in the reaction with oxygen. It rapidly forms boron oxide and water, making it a crucial compound in various chemical syntheses and industrial applications.

Understanding diborane helps chemists develop safer, more efficient chemical processes that harness its powerful reactivity while mitigating its risks.