Problem 97 - Chemistry

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?

Step-by-Step Solution

Answer

The total volume of gases collected from 2.6 x 10^2 g of Nitroglycerin under the given conditions is \(2.03 \times 10^3L\). The partial pressures of the gases can be calculated by dividing the number of moles of each gas by the total moles of gases and then multiplying by the total pressure of 1.2 atm.

Step 1: Calculate Moles of Nitroglycerin

First, we need to calculate the moles of Nitroglycerin used in the chemical reaction. The molar mass of Nitroglycerin: \(C_3H_5(NO_3)_3\) is 227 g/mol. We calculate the moles by taking the mass of Nitroglycerin which is 2.6 x 10^2 g and divide it by its molar mass.\n\nMoles = Mass/Molar Mass = \(2.6 \times 10^2\) g / 227 g/mol = \(1.15 \times 10^1\) mol

Step 2: Calculate Total Moles of Gases

According to the balanced chemical equation, 4 moles of Nitroglycerin decompose to produce 12 moles of CO2, 10 moles of H2O, 6 moles of N2 and 1 mole of O2. It’s a total of 29 moles of gases per 4 moles of Nitroglycerin.\n\nSo total moles of gases from 1.15 x 10^1 mol of Nitroglycerin = \((1.15 \times 10^1)\) mol x \((29/4)\) = \(8.34 \times 10^1\) mol

Step 3: Calculate Total Volume of Gases

We can use Ideal Gas Law to find volume. The ideal gas law equation is \(PV=nRT\). Here, we know the pressure P=1.2atm, Gas constant R=0.0821 \(L.atm/mol.K\), and temperature T is \(25^{\circ} C = 298K\). Plugging these values into the equation, we isolate V (Volume). Volume V = \(nRT/P\) = \((8.34 \times 10^1) mol * 0.0821 L.atm/mol.K * 298K / 1.2 atm = 2.03 \times 10^3L\)

Step 4: Calculate Partial Pressures

The partial pressure of a gas is proportional to its mole fraction. Since we know the moles of each gas, we can find each gases' mole fraction by dividing by the total moles of gas. The partial pressure is each gases' mole fraction times the total pressure of \(1.2 atm\).

Key Concepts

Nitroglycerin Decomposition

Nitroglycerin is well-known for its explosive properties, and it decomposes into several gaseous products. This decomposition reaction is represented by the chemical equation: \[4 \mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(s) \rightarrow 12 \mathrm{CO}_{2}(g) + 10 \mathrm{H}_{2} \mathrm{O}(g) + 6\mathrm{~N}_{2}(g) + \mathrm{O}_{2}(g)\]Understanding this reaction is crucial when exploring chemical energetics involving nitroglycerin.

• Reactants and Products: The decomposition begins with 4 moles of nitroglycerin, resulting in 29 moles of gaseous products: carbon dioxide \((\mathrm{CO}_{2})\), water vapor \((\mathrm{H}_{2}\mathrm{O})\), nitrogen gas \((\mathrm{N}_{2})\), and oxygen gas \((\mathrm{O}_{2})\).
• Energy Release: Besides producing gases, nitroglycerin decomposition releases a large amount of energy, amplifying its explosive power.

The rapid decomposition into gases and the associated energy release make nitroglycerin a potent explosive, frequently used in the construction and demolition industries.

Chemical Reaction Stoichiometry

Stoichiometry is a vital concept that allows us to relate the quantities of reactants and products in a chemical reaction. Based on the balanced chemical equation for nitroglycerin, we use stoichiometry to determine the amounts of gases formed.

• Mole Ratios: This equation shows that decomposing 4 moles of nitroglycerin gives 12 moles of \(\mathrm{CO}_{2}\), 10 moles of \(\mathrm{H}_{2}\mathrm{O}\), 6 moles of \(\mathrm{N}_{2}\), and 1 mole of \(\mathrm{O}_{2}\). These ratios are central for calculating the product formation.
• Calculating Total Gases: For the initial 11.5 moles of nitroglycerin used in the exercise, the total moles of gases can be calculated using the ratio \((29/4)\): \[ \text{Total moles of gases} = 11.5 \text{ moles} \times \frac{29}{4} = 83.4 \text{ moles}\]

Understanding stoichiometry allows us to efficiently transition from known reactant quantities to predicting products, a foundational skill in chemistry.

Partial Pressure Calculation

In a mixture of gases, the pressure exerted by each type of gas is referred to as its partial pressure. Dalton's Law states that the total pressure exerted by a mixture is the sum of the partial pressures of individual gases. This concept is applied in situations like the nitroglycerin decomposition exercise.

• Mole Fraction: First, determine each gas's mole fraction, which is the number of moles of the specific gas divided by the total moles of gas:
\[\text{Mole fraction for a gas} = \frac{\text{Moles of the gas}}{\text{Total moles of gases}}\]
• Calculating Partial Pressure: Then, the partial pressure can be found using the relation:\[\text{Partial Pressure} = \text{Mole Fraction} \times \text{Total Pressure}\]
• Example:If a component gas has a mole fraction of 0.4 in a 1.2 atm system, its partial pressure is 0.48 atm \((0.4 \times 1.2)\).

This method neatly analyzes atmospheric conditions or situations involving gas mixtures. By understanding and applying partial pressure calculations, one can accurately predict individual gas behaviors in a mixture.