Problem 106 - Chemistry The Central Science In Si Units

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?

Step-by-Step Solution

Answer

(a) \( \mathrm{H}_3\mathrm{BO}_3 + \mathrm{NH}_3 \rightarrow \mathrm{BN} + 3\mathrm{H}_2\mathrm{O} \). (b) 90.31 g of \( \mathrm{BN} \) produced. (c) 5.17 moles of \( \mathrm{NH}_3 \) leftover. (d) Film area is 1075 \( \mathrm{cm}^2 \).

Step 1: Write the balanced chemical equation

To solve this, start by identifying the chemical reaction. The reactants are boric acid \( \mathrm{H}_3\mathrm{BO}_3 \) and ammonia \( \mathrm{NH}_3 \), which form boron nitride \( \mathrm{BN} \) and water \( \mathrm{H}_2\mathrm{O} \). Balance the equation: \[ \mathrm{H}_3\mathrm{BO}_3 + \mathrm{NH}_3 \rightarrow \mathrm{BN} + 3\mathrm{H}_2\mathrm{O} \] This equation is balanced in terms of atoms for each element.

Step 2: Calculate moles of reactants

To find the potential mass of \( \mathrm{BN} \), calculate the moles of each reactant. First, find the molar mass of \( \mathrm{H}_3\mathrm{BO}_3 \), which is approximately \( 61.83 \mathrm{~g/mol} \), and for \( \mathrm{NH}_3 \), it is \( 17.03 \mathrm{~g/mol} \). - Moles of \( \mathrm{H}_3\mathrm{BO}_3 = \frac{225 \mathrm{~g}}{61.83 \mathrm{~g/mol}} \approx 3.64 \mathrm{~mol} \).- Moles of \( \mathrm{NH}_3 = \frac{150 \mathrm{~g}}{17.03 \mathrm{~g/mol}} \approx 8.81 \mathrm{~mol} \).

Step 3: Determine the limiting reactant

In the balanced equation, the molar ratio of \( \mathrm{H}_3\mathrm{BO}_3 \) to \( \mathrm{BN} \) is 1:1. Thus, \( \mathrm{H}_3\mathrm{BO}_3 \) is the limiting reactant because it has fewer moles \( (3.64 \text{ moles}) \) compared to \( \mathrm{NH}_3 \).

Step 4: Calculate the mass of \( \mathrm{BN} \) produced

Using the moles of the limiting reactant \( \mathrm{H}_3\mathrm{BO}_3 \), calculate the mass of boron nitride produced. Molar mass of \( \mathrm{BN} \) is \( 24.82 \mathrm{~g/mol} \). The mass of \( \mathrm{BN} \) is \[ 3.64 \mathrm{~mol} \times 24.82 \mathrm{~g/mol} = 90.31 \mathrm{~g} \]

Step 5: Calculate leftover reactant

Subtract the moles of \( \mathrm{NH}_3 \) that reacted \( (3.64 \text{ moles}) \) from the initial moles \( 8.81 \). Leftover moles of \( \mathrm{NH}_3 \) are \[ 8.81 - 3.64 = 5.17 \mathrm{~mol} \]

Step 6: Calculate film area from BN mass

Given a film thickness of \( 0.4 \mathrm{~mm} = 0.04 \mathrm{~cm} \), use the BN density \( 2.1 \mathrm{~g/cm}^3 \) and mass \( 90.31 \mathrm{~g} \) to find the volume, and then the area:\[ \text{Volume} = \frac{90.31 \mathrm{~g}}{2.1 \mathrm{~g/cm}^3} = 43.00 \mathrm{~cm}^3 \] Area: \[ \text{Area} = \frac{43.00 \mathrm{~cm}^3}{0.04 \mathrm{~cm}} = 1075 \mathrm{~cm}^2 \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's like a recipe that tells you how much of each ingredient you need to make a certain amount of product. The idea is to use balanced chemical equations to calculate the amounts of products and reactants. In our exercise, stoichiometry helps to determine how much boron nitride (\( \mathrm{BN} \) can be synthesized from given amounts of boric acid and ammonia.

• We begin by writing and balancing the chemical equation. It's crucial as it provides the necessary ratios of reactants and products.
• Calculate moles of each reactant using their respective molar masses. Moles provide a common measure for comparing chemical amounts.
• Use these moles, in accordance with the balanced equation, to understand the reaction's progression and outputs.

Stoichiometry makes sure the reaction's proportions are respected, ensuring accurate and efficient use of materials.

Limiting Reactant

Determining the limiting reactant is a key aspect of stoichiometry. The limiting reactant is the substance that will run out first, thus limiting the amount of product that can form. In our example, boric acid (\( \mathrm{H}_3\mathrm{BO}_3 \) acts as the limiting reactant.

• We first calculate the moles of each reactant. This involves dividing the mass of each reactant by its molar mass.
• Compare these moles to the stoichiometric ratios provided by the balanced chemical equation.
• The reactant with the smallest stoichiometric ratio is the limiting reactant.

Once identified, the limiting reactant governs the maximum amount of product (boron nitride) that can be produced. This identification is crucial as it allows chemists to predict yields and efficiently use resources.

Molar Mass

Molar mass is a vital concept in chemistry. It refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It acts as a bridge between the macroscopic and microscopic worlds of chemical substances.

• Molar mass of a compound is calculated by adding up the atomic masses of its constituent elements.
• In this exercise, molar mass allows us to convert between grams and moles for reactants and products.

For example:

• \( \mathrm{H}_3\mathrm{BO}_3 \) has a molar mass of approximately 61.83 g/mol.
• \( \mathrm{NH}_3 \) has a molar mass of about 17.03 g/mol.
• \( \mathrm{BN} \) has a molar mass of 24.82 g/mol.

By using molar masses, we can navigate through the problem, converting masses to moles or vice versa as needed, allowing for precise calculations.

Balanced Chemical Equation

A balanced chemical equation is like a recipe that ensures ingredients are combined in the correct proportions to produce the final dish without leftovers. It is crucial for stoichiometry calculations because it provides the ratios needed to relate reactants to products.Once the equation is balanced, it conveys important information such as:

• The type of reactants and products involved in the chemical reaction.
• The molecular or molar ratios of these compounds.
• The conservation of mass, as the number of each type of atom remains constant on both sides of the equation.

The balanced equation used in the exercise is:\[ \mathrm{H}_3\mathrm{BO}_3 + \mathrm{NH}_3 \rightarrow \mathrm{BN} + 3 \mathrm{H}_2\mathrm{O} \]Here, each component’s fully balanced, reflecting the reaction’s stoichiometry. Balancing ensures all reactants and products are in the correct proportions, providing a clear roadmap to solve quantitative problems.