Question

Diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\), is valuable for the synthesis of new organic compounds. The boron compound can be made by the reaction $$2 \mathrm{NaBH}_{4}(\mathrm{~s})+\mathrm{I}_{2}(\mathrm{~s}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+2 \mathrm{NaI}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g})$$ Suppose you use \(1.203 \mathrm{~g} \mathrm{NaBH}_{4}\) and excess iodine, and you isolate \(0.295 \mathrm{~g} \mathrm{~B}_{2} \mathrm{H}_{6} .\) Calculate the percent yield of \(\mathrm{B}_{2} \mathrm{H}_{6}\)

Short answer

The percent yield of \(\mathrm{B}_{2} \mathrm{H}_{6}\) is 67.05\%.

Step-by-step solution

Calculate Molar Mass

First, we need to find the molar masses of the reactants and products involved in the reaction. The molar mass of \(\mathrm{NaBH}_{4}\) is: \(\mathrm{Na} = 22.99\), \(\mathrm{B} = 10.81\), \(\mathrm{H} = 1.01 \times 4 = 4.04\). So, the molar mass is: \[22.99 + 10.81 + 4.04 = 37.84\text{ g/mol}\] The molar mass of \(\mathrm{B}_{2} \mathrm{H}_{6}\) is: \(\mathrm{B} = 10.81 \times 2 = 21.62\), \(\mathrm{H} = 1.01 \times 6 = 6.06\). So, \[\mathrm{B}_{2} \mathrm{H}_{6} = 21.62 + 6.06 = 27.68\text{ g/mol}\]

Calculate Moles of Sodium Borohydride

Use the mass and molar mass to find the moles of \(\mathrm{NaBH}_{4}\): \[\text{Moles of } \mathrm{NaBH}_{4} = \frac{1.203\text{ g}}{37.84\text{ g/mol}} \approx 0.0318\text{ mol}\]

Use Stoichiometry to Find Theoretical Yield

From the balanced equation, 2 moles of \(\mathrm{NaBH}_{4}\) produce 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\). Thus, \[\text{Theoretical moles of } \mathrm{B}_{2} \mathrm{H}_{6} = \frac{0.0318\text{ mol}}{2} \approx 0.0159\text{ mol}\] Now, convert moles of \(\mathrm{B}_{2} \mathrm{H}_{6}\) to grams: \[0.0159\text{ mol} \times 27.68\text{ g/mol} = 0.440\text{ g} \] This is the theoretical yield of \(\mathrm{B}_{2} \mathrm{H}_{6}\).

Calculate Percent Yield

Percent yield is determined using the formula: \[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]Substitute the values: \[\text{Percent Yield} = \left( \frac{0.295\text{ g}}{0.440\text{ g}} \right) \times 100\% = 67.05\%\]

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative relationship between reactants and products in a chemical reaction. It's essentially the toolset for figuring out how much of each substance is needed or produced.
For the chemical reaction involving sodium borohydride (\(\mathrm{NaBH}_{4}\)) and iodine (\(\mathrm{I}_{2}\)), we use stoichiometry to predict the amount of diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)) that can be produced. This involves:

• Understanding the balanced chemical equation: \[2 \mathrm{NaBH}_{4}(\mathrm{s}) + \mathrm{I}_{2}(\mathrm{s}) \rightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g}) + 2 \mathrm{NaI}(\mathrm{s}) + \mathrm{H}_{2}(\mathrm{g})\]
• Identifying the molar ratios between the reactants and products from the equation. In this case, 2 moles of \(\mathrm{NaBH}_{4}\) produce 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\).

By applying stoichiometry, we can calculate the theoretical amount of product formed (theoretical yield), given a certain amount of reactant. This sets the stage for further calculations, such as determining the percent yield.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance and is expressed in grams per mole. Calculating molar mass is crucial for converting between moles and grams, a key step in stoichiometric calculations.
To calculate the molar mass of a compound, like \(\mathrm{NaBH}_{4}\), you simply add together the atomic masses of all the atoms in the formula:

• \(\mathrm{NaBH}_{4}\): \(\mathrm{Na} = 22.99\), \(\mathrm{B} = 10.81\), \(\mathrm{H} = 1.01\times4 = 4.04\), which sums up to \(37.84\text{ g/mol}\).
• For \(\mathrm{B}_{2} \mathrm{H}_{6}\), calculate similarly: \(\mathrm{B} = 10.81\times2 = 21.62\), \(\mathrm{H} = 1.01\times6 = 6.06\), totaling \(27.68\text{ g/mol}\).

These calculations allow you to relate the mass of the reactants to the amount in moles, which we use in stoichiometry to predict the amount of product formed.

Theoretical Yield

Theoretical yield refers to the maximum amount of product that can be generated from a given amount of reactants, assuming complete conversion under ideal conditions. It is derived using molar masses and stoichiometry-based mole-to-mole ratios from the balanced chemical equation.
For example, given the reaction between \(\mathrm{NaBH}_{4}\) and \(\mathrm{I}_{2}\), if 1.203 grams of \(\mathrm{NaBH}_{4}\) are used, the theoretical moles of \(\mathrm{B}_{2} \mathrm{H}_{6}\) can be calculated as:

• Convert grams to moles of \(\mathrm{NaBH}_{4}\) using its molar mass: \(\frac{1.203\text{ g}}{37.84\text{ g/mol}} \approx 0.0318\text{ mol}\).
• From stoichiometry, the ratio tells that 2 moles of \(\mathrm{NaBH}_{4}\) produce 1 mole of \(\mathrm{B}_{2} \mathrm{H}_{6}\), so \(\frac{0.0318}{2} \approx 0.0159\text{ mol}\) of \(\mathrm{B}_{2} \mathrm{H}_{6}\) can be formed.
• Convert these moles back to grams using the molar mass of \(\mathrm{B}_{2} \mathrm{H}_{6}\): \(0.0159\times27.68 = 0.440\text{ g}\).

This 0.440 grams is the theoretical yield. Comparing this to the actual yield allows us to determine the efficiency of the reaction through percent yield calculations.