Problem 70 - Chemistry The Central Science In Si Units

Question

Detonation of nitroglycerin proceeds as follows: $$ \begin{aligned} 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}(l) & \longrightarrow \\ & 12 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) If a sample containing \(2.00 \mathrm{~mL}\) of nitroglycerin (density = \(1.592 \mathrm{~g} / \mathrm{mL})\) is detonated, how many moles of gas are produced? (b) If each mole of gas occupies \(55 \mathrm{~L}\) under the conditions of the explosion, how many liters of gas are produced? (c) How many grams of \(\mathrm{N}_{2}\) are produced in the detonation?

Step-by-Step Solution

Answer

0.1015 mol gas, 5.58 L gas, 0.5886 g \( \text{N}_2 \).

Step 1: Calculate Mass of Nitroglycerin

First, we find the mass of the nitroglycerin sample using its volume and density. The formula is: \( \text{mass} = \text{volume} \times \text{density} \). Given, \( \text{volume} = 2.00 \text{ mL} \) and \( \text{density} = 1.592 \text{ g/mL} \), the mass is \( \text{mass} = 2.00 \times 1.592 = 3.184 \text{ g} \).

Step 2: Calculate Moles of Nitroglycerin

Calculate the moles of nitroglycerin using its molar mass. The molar mass of nitroglycerin \( \text{C}_3\text{H}_5\text{N}_3\text{O}_9 \) is \( (3\times12.01) + (5\times1.01) + (3\times14.01) + (9\times16.00) = 227.09 \text{ g/mol} \). The moles are\( \displaystyle \frac{3.184\text{ g}}{227.09\text{ g/mol}} = 0.014 \text{ mol} \).

Step 3: Calculate Moles of Gas Produced

The reaction balances to show that 4 moles of nitroglycerin produce 29 moles of gas. So, calculate moles of gas from the moles of nitroglycerin. \( \displaystyle \frac{29 \text{ mol gas}}{4\text{ mol nitro}} imes 0.014 \text{ mol nitro} = 0.1015 \text{ mol gas} \).

Step 4: Calculate Volume of Gas Produced

Using the information that each mole of gas occupies 55 L, multiply the moles of gas by the volume per mole. \( 0.1015 \text{ mol gas} \times 55 \text{ L/mol} = 5.5825 \text{ L} \).

Step 5: Calculate Grams of \(N_2\) Produced

From the balanced equation, \(6\) moles of \(N_2\) are produced from \(4\) moles of nitroglycerin. Using this stoichiometry and the molar mass of \(N_2\) (28.02 g/mol), calculate grams of \(N_2\): \( \displaystyle \frac{6\text{ mol N}_2}{4\text{ mol nitro}} \times 0.014 \text{ mol nitro} \times 28.02 \text{ g/mol N}_2 = 0.5886 \text{ g N}_2 \).

Key Concepts

Detonation

Detonation is a type of rapid chemical reaction involving a sharp increase in density, temperature, and pressure. Unlike a regular combustion reaction, detonation happens at supersonic speeds, creating shock waves. This process often results in an explosive force. For example, when nitroglycerin detonates, it turns into various gases in just microseconds. The reaction releases an enormous amount of energy quickly, causing an explosion that results in gases such as carbon dioxide, nitrogen, oxygen, and water vapor.
Understanding detonation is crucial in fields like engineering and safety management, especially in industries using explosive materials. It is essential to control detonation processes to prevent accidents and optimize their use in construction, demolition, and manufacturing.

Gas Laws

Gas laws describe the behavior of gases in terms of pressure, volume, and temperature. They help us predict how gases will react under different conditions. One common aspect we refer to in gas laws is the volume occupied by gases. For instance, in the context of detonation, we consider the volume of gas produced from the explosion.
A crucial relationship is the Ideal Gas Law, expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. This equation is useful for calculating unknown conditions of a gas when the other variables are known.
In explosive reactions like nitroglycerin's detonation, knowing the volume each mole of gas occupies under specific conditions helps us calculate the total volume of gases produced. This information can be critical in managing the outcomes of a controlled detonation.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is a vital concept in chemistry that helps predict the amounts of substances consumed and produced in a reaction. The balanced chemical equation of a reaction provides the ratios of different substances involved.
For example, the detonation of nitroglycerin involves stoichiometry to calculate how much gas will be produced from a given amount of reactant. The reaction equation shows us that 4 moles of nitroglycerin produce 29 moles of gas. This information enables us to determine the moles of gas generated from any sample size of nitroglycerin.
Mastering stoichiometry is important for anyone working in chemistry-related fields, as it ensures accurate and safe calculations. It also supports efficient use of resources in various industries.

Mass-to-Mole Conversion

Mass-to-mole conversion is a fundamental concept in chemistry used to relate the mass of a substance to the amount in moles. This conversion uses the molar mass of a substance, which is the mass of one mole of that substance. The molar mass is often found on the periodic table and is measured in grams per mole (g/mol).
To convert mass to moles, divide the given mass by the molar mass. For example, in the case of nitroglycerin, we calculate its molar mass and then use it to find how many moles are in a 3.184 g sample. This conversion is an essential step in determining how much of a reactive substance is present, which affects all subsequent calculations like gas production or yield of products in reactions.
Having a good grasp of mass-to-mole conversion allows chemists to understand reactions stoichiometrically, ensuring precision in experimental and industrial chemical processes.