Question

Nitroglycerin, an explosive compound, decomposes according to the equation \(4 \mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(s) \longrightarrow 12 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) Calculate the total volume of gases when collected at 1.2 atm and \(25^{\circ} \mathrm{C}\) from \(2.6 \times 10^{2} \mathrm{~g}\) of nitroglycerin. What are the partial pressures of the gases under these conditions?

Short answer

Total volume is approximately 678.1 L. Partial pressures: \( \text{CO}_2 \approx 0.496 \) atm, \( \text{H}_2\text{O} \approx 0.414 \) atm, \( \text{N}_2 \approx 0.248 \) atm, \( \text{O}_2 \approx 0.041 \) atm.

Step-by-step solution

Calculate Moles of Nitroglycerin

First, find the molar mass of nitroglycerin, \( \mathrm{C}_3\mathrm{H}_5(\mathrm{NO}_3)_3 \). The approximate molar mass is 227 g/mol. To find the moles of nitroglycerin, divide the given mass by its molar mass: \( \text{moles of nitroglycerin} = \frac{2.6 \times 10^2 \text{ g}}{227 \text{ g/mol}} \approx 1.15 \text{ mol} \).

Use Stoichiometry to Find Moles of Gases Produced

According to the balanced chemical equation, 1 mole of nitroglycerin produces \(12 + 10 + 6 + 1 = 29\) moles of gas. Therefore, \( 1.15 \) moles of nitroglycerin will produce \(1.15 \times 29 \approx 33.35 \) moles of gas.

Apply the Ideal Gas Law to Find Total Volume

Use the ideal gas law \( PV = nRT \) where \( P = 1.2 \text{ atm}, \) \( V \text{ is the volume}, \) \( n = 33.35 \text{ mol}, \) \( R = 0.0821 \text{ L.atm/mol.K}, \) and \( T = 298 \text{ K (which is 25°C + 273)} \). Rearrange to find \( V \): \( V = \frac{nRT}{P} = \frac{33.35 \times 0.0821 \times 298}{1.2} \approx 678.1 \text{ L} \).

Calculate Partial Pressures of Each Gas

Calculate moles of each gas: \(12 \text{ moles of } \text{CO}_2\), \(10 \text{ moles of } \text{H}_2\text{O}\), \(6 \text{ moles of } \text{N}_2\), and \(1 \text{ mole of } \text{O}_2\). Use \( P_i = \frac{n_i}{n_{total}} \times P_{total} \) to find partial pressures: \( P_{\text{CO}_2} = \frac{12}{29} \times 1.2 \approx 0.496 \text{ atm} \), \( P_{\text{H}_2\text{O}} = \frac{10}{29} \times 1.2 \approx 0.414 \text{ atm} \), \( P_{\text{N}_2} = \frac{6}{29} \times 1.2 \approx 0.248 \text{ atm} \), and \( P_{\text{O}_2} = \frac{1}{29} \times 1.2 \approx 0.041 \text{ atm} \).

Key concepts

Nitroglycerin Decomposition

Nitroglycerin is well-known for its explosive properties. When it decomposes, it breaks down into several gases according to the chemical equation provided. For decomposition, each molecule of nitroglycerin can be envisioned as a complex mix of elements just waiting to explode into simpler gases.
Upon decomposition, it releases gases such as carbon dioxide (\(\text{CO}_2\)), water vapor (\(\text{H}_2\text{O}\)), nitrogen (\(\text{N}_2\)), and oxygen (\(\text{O}_2\)). The energy released in this breakdown drives the explosive force of nitroglycerin. This means, every mole of nitroglycerin can unleash a large number of gaseous moles.
To predict the result of this reaction, we first need to understand the breakdown from solid nitroglycerin to these gases and the quantity each reaction produces. This leads naturally to the next key concept: stoichiometry.

Stoichiometry

Stoichiometry might sound complicated, but it's simply the method we use to figure out how much product we get from a certain amount of reactant in a chemical reaction. Let's break it down step by step.
In our reaction, stoichiometry tells us that from each mole of nitroglycerin (C\(_3\)H\(_5\)(NO\(_3\))\(_3\)), 29 moles of gases are formed. To find out the total amount of each gas formed when decomposing 2.6 x 10\(^2\) g of nitroglycerin, we first converted the mass of nitroglycerin to moles by dividing by its molar mass, which is approximately 227 g/mol.

• We calculated approximately 1.15 moles of nitroglycerin from the given mass.
• By using the stoichiometric ratio, 1.15 moles of nitroglycerin would then result in roughly 33.35 moles of gas.

With stoichiometry, we effectively predict the output of our chemical reactions, thereby providing crucial input for our next calculation involving the ideal gas law.

Partial Pressures

Partial pressure is a concept from chemistry that describes the pressure contributed by each type of gas in a mixture. Just like how different people contribute to a group project, each gas component in a mix contributes to the total pressure exerted by that mix of gases.
At the end of the decomposition, we're given the total pressure is 1.2 atm. From stoichiometry, we know how many moles of each gas are involved:

• 12 moles of CO\(_2\)
• 10 moles of H\(_2\)O
• 6 moles of N\(_2\)
• 1 mole of O\(_2\)

To find the partial pressure of each gas, we use the formula: (\( P_i = \frac{n_i}{n_{total}} \times P_{total} \)). This lets us figure out the pressure each gas would exert alone in the mixture, helping us understand how each component behaves in the context of the whole.

Chemical Equations

Understanding chemical equations is essential when dealing with reactions like the decomposition of nitroglycerin. These equations are like a recipe showing the "ingredients" we start with and the "dishes" we end up with.
Our equation starts with 4 moles of the solid nitroglycerin decomposing into 29 moles of various gases. A chemical equation not only shows the substances involved but also their ratios, which is critical in calculations like the volume of gas produced or the partial pressures of individual gases.
To interpret a chemical equation correctly, each formula represents a molecule, and the coefficients (the numbers in front) tell us how many of those molecules participate in the reaction:

• The equation helps us visualize and calculate transformations from reactants into products.
• It balances different atoms on both sides, ensuring matter isn't created or destroyed but merely rearranged.

A well-balanced chemical equation is fundamental not only for understanding the nature of the reaction but also for solving problems related to quantity and concentration of reactants and products.