Question

Boron nitride, \(\mathrm{BN}\), is an electrical insulator with remarkable thermal and chemical stability. Its density is \(2.1 \mathrm{~g} / \mathrm{cm}^{3}\). It can be made by reacting boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3}\), with ammonia. The other product of the reaction is water. (a) Write a balanced chemical equation for the synthesis of BN. (b) If you made \(225 \mathrm{~g}\) of boric acid react with \(150 \mathrm{~g}\) ammonia, what mass of BN could you make? (c) Which reactant, if any, would be left over, and how many moles of leftover reactant would remain? (d) One application of \(\mathrm{BN}\) is as thin film for electrical insulation. If you take the mass of BN from part (a) and make a \(0.4 \mathrm{~mm}\) thin film from it, what area, in \(\mathrm{cm}^{2}\), would it cover?

Short answer

The balanced chemical equation for the synthesis of boron nitride (BN) is: \( 2\ \mathrm{H}_3 \mathrm{BO}_3 + 6\ \mathrm{NH}_3 \rightarrow 2\ \mathrm{BN} + 6\ \mathrm{H}_2\mathrm{O} \). Given the masses of boric acid and ammonia provided, 72.97 g of BN can be produced, with 0.70 moles of boric acid remaining as the leftover reactant. The BN thin film with a thickness of 0.4 mm would cover an area of 868.75 cm².

Step-by-step solution

Write the unbalanced equation

First, write the unbalanced chemical equation for the reaction between boric acid (H3BO3) and ammonia (NH3) to produce boron nitride (BN) and water (H2O). \[ \mathrm{H}_3 \mathrm{BO}_3 + \mathrm{NH}_3 \rightarrow \mathrm{BN} + \mathrm{H}_2\mathrm{O} \]

Balance the equation

Now, we will balance the chemical equation by adjusting the stoichiometric coefficients: \[ 2\ \mathrm{H}_3 \mathrm{BO}_3 + 6\ \mathrm{NH}_3 \rightarrow 2\ \mathrm{BN} + 6\ \mathrm{H}_2\mathrm{O} \] So, the balanced chemical equation is: \( 2\ \mathrm{H}_3 \mathrm{BO}_3 + 6\ \mathrm{NH}_3 \rightarrow 2\ \mathrm{BN} + 6\ \mathrm{H}_2\mathrm{O} \) #b) What mass of BN could be made?#

Determine moles of reactants

Convert grams of boric acid and ammonia to moles using their respective molar masses. Molar mass of H3BO3 = 3(1.01 g/mol) + 1(10.81 g/mol) + 3(16.00 g/mol) = 61.83 g/mol Moles of H3BO3 = \(\frac{225 \ \text{g}}{61.83 \ \mathrm{g/mol}}\) = 3.64 moles Molar mass of NH3 = 1(14.01 g/mol) + 3(1.01 g/mol) = 17.03 g/mol Moles of NH3 = \(\frac{150 \ \text{g}}{17.03 \ \mathrm{g/mol}}\) = 8.81 moles

Identify the limiting reactant and find moles of BN produced

From the balanced equation, it can be seen that the mole ratio of H3BO3 to NH3 is 1:3. Dividing the moles of each reactant by their respective coefficients in the balanced equation, we get: For H3BO3: \(\frac{3.64 \text{ moles}}{2}\) = 1.82 For NH3: \(\frac{8.81 \text{ moles}}{6}\) = 1.47 Since 1.47 < 1.82, ammonia (NH3) is the limiting reactant. From the balanced equation, 6 moles of NH3 will produce 2 moles of BN. Moles of BN produced = \(\frac{2}{6}\times 8.81 \ \text{moles}\) = 2.94 moles

Convert moles of BN to grams

Find the molar mass of BN and convert the moles of BN to grams: Molar mass of BN = 1(10.81 g/mol) + 1(14.01 g/mol) = 24.82 g/mol Mass of BN produced = \( 2.94 \ \text{moles} \times 24.82 \ \mathrm{g/mol}\) = 72.97 g So, 72.97 g of BN can be produced. #c) Which reactant is leftover and how many moles would remain?#

Determine the leftover reactant and remaining moles

Since ammonia was the limiting reactant, the boric acid is the leftover reactant. Calculate the moles of boric acid that reacted with ammonia and subtract it from the initial moles to get the remaining moles of boric acid: Moles of H3BO3 reacted = \(\frac{2}{6}\times 8.81 \ \text{moles}\) = 2.94 moles Moles of H3BO3 remaining = 3.64 moles - 2.94 moles = 0.70 moles So, 0.70 moles of boric acid would remain. #d) What area would the BN thin film cover?#

Convert thickness of the BN film to cm

The thickness of the BN film is given in mm. Convert it to cm: Thickness = 0.4 mm = 0.04 cm

Calculate the volume of BN produced

Use the mass and density of BN to calculate its volume: Volume = \(\frac{\text{mass of BN}}{\text{density}}\) = \(\frac{72.97 \ \text{g}}{2.1 \ \mathrm{g/cm}^3}\) = 34.75 cm³

Calculate area covered by BN film

Use the volume and thickness of the BN film to calculate the area it covers: Area = \(\frac{\text{volume of BN}}{\text{thickness}}\) = \(\frac{34.75 \ \mathrm{cm}^3}{0.04 \ \mathrm{cm}}\) = 868.75 cm² So, the area covered by the BN thin film is 868.75 cm².

Key concepts

Limiting Reactant

In any chemical reaction, reactants are combined to form products. However, often, one reactant is completely consumed before the others. This reactant is known as the limiting reactant. It determines the maximum amount of product that can be formed, as the reaction will stop when this reactant is used up entirely. In the given problem, the reaction uses boric acid and ammonia to produce boron nitride and water. To find the limiting reactant, one must:

• Determine the molecular ratio from the balanced equation, which tells how many moles of each reactant are needed.
• Calculate the number of moles of each reactant present in the reaction.
• Divide the moles of each reactant by its stoichiometric coefficient from the balanced equation.
• The reactant with the smallest resulting value is the limiting reactant.

In this particular reaction, ammonia is the limiting reactant because it can produce the least amount of product when compared to boric acid.

Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. By using the balanced chemical equation, you can predict the amount of products formed from given quantities of reactants, or vice-versa. The stoichiometric coefficients from the balanced equation establish the proportions in which reactants combine and products form. For the reaction of boric acid with ammonia to form boron nitride, stoichiometry allows us to:

• Quantify the exact amount of ammonia needed to completely react with a given amount of boric acid, based on a 1:3 mole ratio from the balanced equation.
• Calculate the exact amount of product formed from these reactants.

This method not only lets us predict product amounts but also helps minimize waste by determining which reactant will run out first.

Mole Calculations

Moles are a fundamental unit in chemistry that allow us to relate mass to the number of particles. Calculations involving moles are essential to convert between the mass of a substance and the amount of chemical substance. Here's how mole calculations apply:

• Firstly, find the molar masses of the reactants and products, which are the mass per mole of that compound, typically in g/mol.
• Convert the masses of boric acid and ammonia to moles using their respective molar masses, which helps in determining the amount available for reaction.
• Use the mole quantities and the stoichiometry of the reaction to find out how much product, in moles and mass, is formed from these reactants.

Mole calculations allow for an accurate understanding of how much substance is needed or will result from chemical reactions.

Balanced Chemical Equation

A balanced chemical equation ensures that there is the same number of each type of atom on both the reactant and product sides. This complies with the Law of Conservation of Mass, indicating that matter cannot be created or destroyed in a chemical reaction. Here's how the balancing process was executed:

• Initially, write an unbalanced equation based on the description of the reaction—here, between boric acid and ammonia forming boron nitride and water.
• Count the number of each type of atom on both sides of the equation.
• Adjust the coefficients to obtain the same number of each type of atom on both sides. For instance, balancing hydrogen atoms with coefficients results in a proper equation.

A balanced chemical equation not only predicts the quantity of products but also gives insight into the reactant quantities needed for the reaction.