91Ó°ÊÓ

One way to synthesize diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\), is the reaction \(\begin{aligned} 2 \mathrm{NaBH}_{4}(\mathrm{s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(\ell) & \rightarrow \\ \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+& 2 \mathrm{NaH}_{2} \mathrm{PO}_{4}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \end{aligned}\) (a) If you have \(0.136 \mathrm{g}\) of \(\mathrm{NaBH}_{4}\) and excess \(\mathrm{H}_{3} \mathrm{PO}_{4},\) and you collect the resulting \(\mathrm{B}_{2} \mathrm{H}_{6}\) in a 2.75 -L flask at \(25^{\circ} \mathrm{C},\) what is the pressure of the \(\mathrm{B}_{2} \mathrm{H}_{6}\) in the flask? (b) A by-product of the reaction is \(\mathrm{H}_{2}\) gas. If both \(\mathrm{B}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2}\) gas come from this reaction, what is the total pressure in the \(2.75-\mathrm{L}\) flask (after reaction of 0.136 g of NaBH_with excess \(\left.\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) at \(25^{\circ} \mathrm{C} ?\)

Step by step solution

01

Determine Molar Mass of Reactants

Find the molar mass of NaBH_4. Na = 22.99 g/mol, B = 10.81 g/mol, H = 1.01 g/mol. Thus, the molar mass of NaBH_4 is 22.99 + 10.81 + (4 * 1.01) = 37.84 g/mol.

02

Calculate Moles of NaBH_4

Use the mass given to find moles of NaBH_4: \[ \text{Moles of } \mathrm{NaBH}_{4} = \frac{0.136 \text{ g}}{37.84 \text{ g/mol}} = 0.0036 \text{ mol} \]

03

Use Stoichiometry to Find Moles of B_2H_6

From the equation, 2 moles of NaBH_4 produce 1 mole of B_2H_6. So, 0.0036 moles of NaBH_4 will produce: \[ \frac{0.0036 \text{ mol NaBH}_4}{2} = 0.0018 \text{ mol B}_2H_6 \]

04

Calculate Pressure of B_2H_6 Using Ideal Gas Law

Use PV = nRT to find pressure. R = 0.0821 L atm / K mol, T = 298 K (25°C + 273). \[ P \times 2.75 = 0.0018 \times 0.0821 \times 298 \] \[ P = \frac{0.0018 \times 0.0821 \times 298}{2.75} \approx 0.0161 \text{ atm} \]

05

Find Moles of H_2 Produced

From the balanced equation, 2 moles of NaBH_4 produce 2 moles of H_2 gas. So, 0.0036 moles of NaBH_4 will produce 0.0036 moles of H_2.

06

Calculate Pressure of H_2 Using Ideal Gas Law

Using PV = nRT again for H_2: \[ P \times 2.75 = 0.0036 \times 0.0821 \times 298 \] \[ P = \frac{0.0036 \times 0.0821 \times 298}{2.75} \approx 0.0323 \text{ atm} \]

07

Calculate Total Pressure in the Flask

The total pressure is the sum of the pressures of B_2H_6 and H_2. \[ P_{\text{total}} = 0.0161 + 0.0323 = 0.0484 \text{ atm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant, \( 0.0821 \text{ L atm / K mol} \).
• \( T \) is the temperature in Kelvin.

This equation helps us predict how gases behave under different conditions. It's particularly useful for calculating pressures of gases after a chemical reaction.
For instance, in the reaction generating diborane and hydrogen gas, we use the Ideal Gas Law to find the pressure of each gas in a flask. By knowing the moles of gas and the flask's conditions, you can easily find the pressure exerted by the gas molecules.

Calculating Molar Mass

Molar mass is an essential concept for understanding amounts in chemistry. It is the mass of one mole of a substance, usually given in g/mol.
To calculate the molar mass of a compound, sum the atomic masses of its elements. For example, sodium borohydride (\(\text{NaBH}_4\)) has a molar mass calculated as:

• Sodium (Na): 22.99 g/mol
• Boron (B): 10.81 g/mol
• Hydrogen (H): 4 x 1.01 g/mol

Add these values to get 37.84 g/mol. Knowing the molar mass is crucial for converting between mass and moles, a vital step in stoichiometry to link physical measurements to chemical equations.

The Art of Chemical Equation Balancing

Balancing chemical equations ensures that the same number of atoms exists on both sides of the equation. This is pivotal in the conservation of mass principle.
In stoichiometry, balanced equations provide the necessary ratios to relate moles of reactants to moles of products. For example, the equation \(2 \text{NaBH}_4 + 2 \text{H}_3\text{PO}_4 \rightarrow \text{B}_2\text{H}_6 + 2 \text{NaH}_2\text{PO}_4 + 2 \text{H}_2\) tells us that:

• 2 moles of \(\text{NaBH}_4\) yield 1 mole of \(\text{B}_2\text{H}_6\)
• 2 moles of \(\text{NaBH}_4\) also yield 2 moles of \(\text{H}_2\)

Balanced equations are a roadmap for any chemical calculation, ensuring that all stoichiometric relationships are respected and maintained.