Question

Detonation of nitroglycerin proceeds as follows: \(4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}(l)\) $$ 12 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) If a sample containing \(2.00 \mathrm{~mL}\) of nitroglycerin (density \(=1.592 \mathrm{~g} / \mathrm{mL}\) ) is detonated, how many total 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?

Short answer

(a) The total number of moles of gas produced is approximately 0.1015 mol. (b) The volume of the gas produced is approximately 5.58 L. (c) The mass of nitrogen gas produced is approximately 0.588 g.

Step-by-step solution

(a) Finding the moles of nitroglycerin

To find the moles of nitroglycerin in the sample, first, we need to find the mass of the nitroglycerin sample. We can do this using the density formula: mass = density × volume. Density = 1.592 g/mL Volume = 2.00 mL Mass = 1.592 g/mL × 2.00 mL = 3.184 g Now, we need to find the molar mass of nitroglycerin: Molar mass of nitroglycerin (C₃H₅N₃O₉) = 3(12.01 g/mol) + 5(1.01 g/mol) + 3(14.01 g/mol) + 9(16.00 g/mol) = 227.09 g/mol Next, we can find the moles of nitroglycerin by dividing the mass by its molar mass: Moles of nitroglycerin = 3.184 g ÷ 227.09 g/mol ≈ 0.0140 mol

(a) Finding the total moles of gas produced

From the balanced chemical equation, notice the stoichiometric ratios of the gases produced to the nitroglycerin: 4 C₃H₅N₃O₉ → 12 CO₂ + 6 N₂ + O₂ + 10 H₂O So, for every 4 moles of nitroglycerin, 29 moles of gas are produced (12 + 6 + 1 + 10 = 29). Now, we can use the stoichiometric ratio (4:29) to find the total moles of gas produced: Total moles of gas = (0.0140 mol nitroglycerin) × (29 moles gas / 4 moles nitroglycerin) ≈ 0.1015 mol

(b) Finding the volume of gas produced

The problem states that each mole of gas occupies 55 L under the conditions of the explosion. Therefore, we can find the volume of the gas produced by multiplying the total moles of gas by the volume per mole: Volume of gas = (0.1015 mol gas) × (55 L/mol) ≈ 5.58 L

(c) Finding the grams of N2 produced

Again, from the balanced chemical equation, we can see that for every 4 moles of nitroglycerin, 6 moles of N₂ are produced. We can use the stoichiometric ratio (4:6) to find the moles of N₂ produced: Moles of N₂ = (0.0140 mol nitroglycerin) × (6 moles N₂ / 4 moles nitroglycerin) ≈ 0.0210 mol Now, we need to find the molar mass of N₂: Molar mass of N₂ = 2(14.01 g/mol) = 28.02 g/mol Finally, we can find the grams of N₂ produced by multiplying the moles of N₂ by its molar mass: Mass of N₂ = (0.0210 mol N₂) × (28.02 g/mol) ≈ 0.588 g To summarize, (a) the total number of moles of gas produced is approximately 0.1015 mol, (b) the volume of the gas produced is approximately 5.58 L, and (c) the mass of nitrogen gas produced is approximately 0.588 g.

Key concepts

Nitroglycerin Detonation

Understanding nitroglycerin detonation is critical when studying chemical reactions and energy release. Nitroglycerin, a highly volatile substance, decomposes rapidly through a process known as detonation. This type of explosion is a chemical reaction producing gases, heat, and pressure.

The balanced chemical equation for the detonation of nitroglycerin (C3H5N3O9(l)) shows the conversion of the liquid compound into several gas products, including carbon dioxide (CO2), nitrogen (N2), oxygen (O2), and water (H2O) in gaseous form. It’s important to recognize that in stoichiometric calculations, the coefficients of the chemical equation are used to understand the proportions in which reactants convert to products.

For every 4 moles of nitroglycerin that detonate, 29 moles of gas are produced. This ratio is the foundation for further calculations, like determining the total volume of gas produced in an explosion. A clear understanding of this ratio is essential for students to successfully analyze the outcomes of the reaction.

Molar Mass Calculation

The concept of molar mass is integral to stoichiometry and facilitates the conversion between mass and moles of a substance. Molar mass is defined as the mass of one mole of a chemical species and is expressed in grams per mole (g/mol).

Calculating the molar mass of nitroglycerin is a straightforward process but requires careful addition of the atomic masses of each element in the compound. A mole of nitroglycerin consists of carbon (C), hydrogen (H), nitrogen (N), and oxygen (O) atoms. By summing the products of the number of each type of atom by its respective atomic mass, one finds the molar mass of the compound.

• Molar Mass of C3H5N3O9 = 3(12.01 g/mol) + 5(1.01 g/mol) + 3(14.01 g/mol) + 9(16.00 g/mol) = 227.09 g/mol

After calculating the molar mass, it’s used to determine the number of moles present in a given mass of the substance. This step is central for solving stoichiometry problems involving mass-to-mole or mole-to-mass conversions.

Gas Volume Calculation

Upon understanding molar mass, it's imperative to grasp the concept of gas volume calculation in reaction stoichiometry. The volume of gases produced in a reaction can be predicted using the ideal gas law under certain conditions, but sometimes an explosion or reaction provides unique conditions.

In the given problem, the explosion conditions dictate that each mole of gas occupies 55 L. This data is a deviation from standard temperature and pressure conditions, highlighting the need for flexibility in applying stoichiometry concepts to real-world problems. Using the total moles of gases produced, one can calculate the total volume by multiplying by the given volume per mole.

This calculation is useful in contexts such as engineering, environmental science, and safety planning, where understanding the amount of gas released in an event like a nitroglycerin detonation is critical. Students should note these kinds of scenario-specific details, as they are pivotal in translating theoretical principles into practical applications.