Question

Sodium borohydride, \(\mathrm{NaBH}_{4}\), a substance used in the synthesis of many pharmaceutical agents, can be prepared by reaction of NaH with \(\mathrm{B}_{2} \mathrm{H}_{6}\) according to the equation \(2 \mathrm{NaH}+\mathrm{B}_{2} \mathrm{H}_{6} \longrightarrow 2 \mathrm{NaBH}_{4}\) (a) How many grams of \(\mathrm{NaBH}_{4}\) can be prepared by reaction between \(8.55 \mathrm{~g}\) of \(\mathrm{NaH}\) and \(6.75 \mathrm{~g}\) of \(\mathrm{B}_{2} \mathrm{H}_{6} ?\) (b) Which reactant is limiting, and how many grams of the excess reactant will be left over?

Short answer

(a) 13.46 g of NaBH4 can be formed. (b) NaH is the limiting reactant; 1.83 g of B2H6 is left over.

Step-by-step solution

Calculate Molar Masses

First, we need to find the molar mass of each compound involved in the reaction. - Molar mass of NaH: Na = 23.0 g/mol and H = 1.0 g/mol \( \Rightarrow \text{NaH} = 23.0 + 1.0 = 24.0 \, \text{g/mol} \)- Molar mass of \(\text{B}_2\text{H}_6\): B = 10.8 g/mol \( \Rightarrow \text{B}_2\text{H}_6 = 2(10.8) + 6(1.0) = 27.6 \, \text{g/mol} \)- Molar mass of \(\text{NaBH}_4\): B = 10.8 g/mol, H = 1.0 g/mol \( \Rightarrow \text{NaBH}_4 = 23.0 + 10.8 + 4(1.0) = 37.8 \, \text{g/mol} \)

Determine Moles of Reactants

Calculate moles for each reactant from given masses:- Moles of NaH: \( \frac{8.55 \, \text{g}}{24.0 \, \text{g/mol}} = 0.356 \text{ mol} \)- Moles of \(\text{B}_2\text{H}_6\): \( \frac{6.75 \, \text{g}}{27.6 \, \text{g/mol}} = 0.245 \text{ mol} \)

Identify the Limiting Reactant

The balanced equation is \( 2 \text{NaH} + \text{B}_2\text{H}_6 \rightarrow 2 \text{NaBH}_4 \). The stoichiometry ratio is 2 moles of NaH to 1 mole of \(\text{B}_2\text{H}_6\). - Check NaH: \( 0.356 \text{ mol NaH} \cdot \frac{1}{2} = 0.178 \text{ mol } \text{B}_2\text{H}_6 \text{ needed} \)Since we have 0.245 mol of \(\text{B}_2\text{H}_6\), NaH is the limiting reactant as it requires more than available \(\text{B}_2\text{H}_6\).

Calculate Grams of NaBH4 Formed

Using the moles of the limiting reactant (NaH), calculate moles of \(\text{NaBH}_4\):- From the balanced equation, 2 moles of NaH produce 2 moles of \(\text{NaBH}_4\). So, 0.356 moles of NaH produce 0.356 moles of \(\text{NaBH}_4\).- Grams of \(\text{NaBH}_4\) produced: \( 0.356 \text{ mol} \times 37.8 \, \text{g/mol} = 13.46 \, \text{g} \)

Calculate Amount of Excess Reactant

Determine how much of \(\text{B}_2\text{H}_6\) was used:- From Step 3, we know that 0.178 moles of \(\text{B}_2\text{H}_6\) are actually used.- Grams of \(\text{B}_2\text{H}_6\) used: \( 0.178 \, \text{mol} \times 27.6 \, \text{g/mol} = 4.92 \, \text{g} \)- Excess \(\text{B}_2\text{H}_6\): \( 6.75 \, \text{g} - 4.92 \, \text{g} = 1.83 \, \text{g} \) of \(\text{B}_2\text{H}_6\) leftover.

Key concepts

Limiting Reactant

In a chemical reaction, a limiting reactant is the substance that is completely consumed first, stopping the reaction from continuing because no additional product can be generated without it. This concept is crucial for predicting the quantities of products formed in the reaction.

Let's illustrate this with the reaction of sodium borohydride (\(\text{NaBH}_4\)) from sodium hydride (\(\text{NaH}\)) and diborane (\(\text{B}_2\text{H}_6\)). Based on the balanced equation:- 2 moles of NaH react with 1 mole of \(\text{B}_2\text{H}_6\) to form 2 moles of \(\text{NaBH}_4\).

In our exercise, we calculated:

• 0.356 moles of NaH
• 0.245 moles of \(\text{B}_2\text{H}_6\)

When doing stoichiometry, you want to determine which of these moles will run out first. For NaH, 0.356 moles require 0.178 moles of \(\text{B}_2\text{H}_6\), less than what's available. Therefore, NaH is used up completely making it the limiting reactant.

Once you identify the limiting reactant, it becomes straightforward to calculate the maximum product yield, as the amount of the product is directly linked to the limiting reactant quantity.

Molar Mass Calculation

To perform stoichiometry calculations, you need to first calculate the molar masses of each of the substances involved. This allows you to convert between grams and moles, which is essential for stoichiometric analyses.

For the current scenario:

• Molar mass of \(\text{NaH}\) is obtained by adding the atomic masses of Sodium (Na) and Hydrogen (H), which results in 24.0 g/mol.
• \(\text{B}_2\text{H}_6\) has a molar mass calculated by adding the masses of Boron (B) and Hydrogen (H), resulting in 27.6 g/mol.
• Lastly, \(\text{NaBH}_4\) is a result of 37.8 g/mol, combining the masses of Na, B, and H.

Having these values allows anyone performing the calculation to accurately convert the mass of reactants or products into moles, crucial for the stoichiometric calculations needed to solve the problem. Without this step, you can't accurately determine which reactant is limiting or how much product is formed.

Excess Reactant

The excess reactant is the substance that remains after the reaction has completed because it was not consumed entirely. Identifying the excess reactant can help in calculating how much of a reactant remains, which can be crucial for recycling in industrial processes or knowing leftover quantities in laboratory settings.

In the example reaction, once we've determined NaH as the limiting reactant, we can calculate the \(\text{B}_2\text{H}_6\) left over:- We initially have 0.245 moles of \(\text{B}_2\text{H}_6\).- Based on stoichiometry, only 0.178 moles of \(\text{B}_2\text{H}_6\) are consumed. This leaves an excess.

• The calculation shows 1.83 g of \(\text{B}_2\text{H}_6\) remains unreacted.

Knowing the excess helps in both practical applications like cost analysis and efficiency checks, as well as educational exercises when learning reaction stoichiometry.