Question

A \(0.05-\mathrm{g}\) sample of the boron hydride, \(\mathrm{B}_{4} \mathrm{H}_{10},\) is burned in pure oxygen to give \(\mathrm{B}_{2} \mathrm{O}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). $$2 \mathrm{~B}_{4} \mathrm{H}_{10}(\mathrm{~s})+11 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{~B}_{2} \mathrm{O}_{3}(\mathrm{~s})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ Calculate the pressure of the gaseous water in a \(4.25-\mathrm{L}\) flask at \(30 .{ }^{\circ} \mathrm{C}\).

Short answer

The pressure of the gaseous water is approximately 0.0276 atm.

Step-by-step solution

Calculate the moles of B4H10 burned

First, calculate the molar mass of boron hydride \(\mathrm{B}_4\mathrm{H}_{10}\):\[4 \, \times \, 10.81 \, g/mol + 10 \, \times \, 1.008 \, g/mol = 53.32 \, g/mol\]. Next, calculate the moles of \(\mathrm{B}_4\mathrm{H}_{10}\) in the 0.05 g sample:\[\text{moles of } \mathrm{B}_4\mathrm{H}_{10} = \frac{0.05 \text{ g}}{53.32 \text{ g/mol}} \approx 0.0009375 \text{ mol}\].

Determine moles of water produced

From the balanced chemical equation, 2 moles of \(\mathrm{B}_4\mathrm{H}_{10}\) produce 10 moles of water. Calculate moles of water from the moles of \(\mathrm{B}_4\mathrm{H}_{10}\) burned:\[\text{moles of } \mathrm{H}_2\mathrm{O} = \frac{10}{2} \times 0.0009375\approx 0.0046875 \text{ mol}\].

Use Ideal Gas Law to find pressure

Use the ideal gas law \(PV = nRT\) to find the pressure of water vapor. Convert \(30^{\circ}C\) to Kelvin by adding 273.15 (since \(T = 30 + 273.15 = 303.15\ K\)). Using \(R = 0.08206 \ \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1}\), solve for pressure \(P\):\[ P = \frac{nRT}{V} = \frac{0.0046875\text{ mol} \times 0.08206 \text{ L atm/mol K} \times 303.15 \text{ K}}{4.25 \text{ L}} \approx 0.0276 \text{ atm}\].

Key concepts

Chemical Reactions

In a chemical reaction, substances known as reactants are transformed into different substances called products. Each chemical reaction can be expressed through a chemical equation, which provides a clear representation of how the reactants change into the products. In this exercise, the reaction involves burning boron hydride

• Boron hydride ( \(\mathrm{B}_{4} \mathrm{H}_{10}\)) as the reactant
• Pure oxygen ( \(\mathrm{O}_{2}\))
• Resulting in boron oxide ( \(\mathrm{B}_{2} \mathrm{O}_{3}\)) and water ( \(\mathrm{H}_{2} \mathrm{O}\)) as the products.

The balanced chemical equation for this process is:

• \(2 \mathrm{~B}_{4} \mathrm{H}_{10}+11 \mathrm{O}_{2} \rightarrow 4 \mathrm{B}_{2} \mathrm{O}_{3}+10 \mathrm{H}_{2} \mathrm{O}\).

In this equation, the coefficients show the stoichiometric relationships. This relationship indicates the ratio of molecules and moles needed for the reactants to completely react and produce the products. Balancing chemical equations ensures that matter is conserved, as dictated by the law of conservation of mass.
When dealing with chemical reactions, especially those involving gases, understanding these principles is crucial for calculating quantities, including the volume and pressure of gases produced.

Stoichiometry

Stoichiometry is all about the calculation of reactants and products in chemical reactions. It uses the coefficients from balanced equations to determine how much of each substance is consumed or produced.
In our reaction, boron hydride (\(\mathrm{B}_4\mathrm{H}_{10}\)) is burned to produce water (\(\mathrm{H}_2\mathrm{O}\)). The balanced equation tells us that:- 2 moles of \(\mathrm{B}_4\mathrm{H}_{10}\) are needed to produce 10 moles of \(\mathrm{H}_2\mathrm{O}\).
This ratio (2:10) is key to finding out how much water is produced from a given amount of boron hydride.
In step 1 of the solution, we calculated the moles of boron hydride available. Next, step 2 utilizes the stoichiometric ratio to calculate the moles of water produced:

• \[\text{moles of } \mathrm{H}_2\mathrm{O} = \frac{10}{2} \times 0.0009375 \approx 0.0046875 \text{ mol}\]

This calculation is crucial because it leads to how the reaction can further be analyzed to find out other conditions such as gas pressure.

Gas Pressure Calculation

Gas pressure calculation can often be done using the Ideal Gas Law, which is essential for understanding the behavior of gases in reactions. The Ideal Gas Law is expressed as \(PV = nRT\), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume
• \(n\) is the moles of gas
• \(R\) is the ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹)
• \(T\) is the temperature in Kelvin

To find the pressure of the water vapor produced in our chemical reaction, we first determine the temperature in Kelvin, which is \(30^{\circ}C + 273.15 = 303.15 \, K\).
Using the values we know:

• Moles, \(n = 0.0046875\text{ mol}\)
• Temperature, \(T = 303.15\text{ K}\)
• Volume, \(V = 4.25\text{ L}\)

We apply the ideal gas law to solve for pressure:\[ P = \frac{nRT}{V} = \frac{0.0046875\text{ mol} \times 0.08206 \text{ L atm/mol K} \times 303.15 \text{ K}}{4.25 \text{ L}} \approx 0.0276 \text{ atm}\]
This formula allows us to calculate the pressure exerted by the gaseous water, building a comprehensive understanding of how chemical reaction data translates into physical properties of gases.