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?

Short answer

(a) 0.1015 mol of gas is produced. (b) 5.58 L of gas is produced. (c) 0.588 g of N2 is produced.

Step-by-step solution

Calculate moles of nitroglycerin

First, we need to convert the given volume of nitroglycerin (2.00 mL) to mass using the given density (1.592 g/mL). Mass = Volume × Density Mass = 2.00 mL × 1.592 g/mL = 3.184 g Now, we need to find the molar mass of nitroglycerin (C3H5N3O9). Use the atomic masses from the periodic table: C: 12.01 g/mol, H: 1.01 g/mol, N: 14.01 g/mol, O: 16.00 g/mol. Molar mass of nitroglycerin = 3(12.01) + 5(1.01) + 3(14.01) + 9(16.00) = 227.13 g/mol. Now, calculate the moles of nitroglycerin: Moles = Mass/Molar Mass Moles = 3.184 g / 227.13 g/mol = 0.0140 mol

Calculate moles of gas produced

From the balanced equation, we know that 4 moles of nitroglycerin (C3H5N3O9) produce: - 12 moles of CO2, - 6 moles of N2, - 1 mole of O2, and - 10 moles of H2O. Total moles of gas produced = 12 + 6 + 1 + 10 = 29 moles. Now we can use the mole ratio to find the moles of gas produced: Moles of gas produced = moles of nitroglycerin × (moles of gas produced/moles of nitroglycerin) Moles of gas produced = 0.0140 mol × (29 mol/4 mol) = 0.1015 mol

Calculate the volume of gas produced

We are given that each mole of gas occupies 55 L under the conditions of the explosion. To find the volume of gas produced, multiply the moles of gas produced by 55 L/mol: Volume of gas = Moles of gas × Volume per mole Volume of gas = 0.1015 mol × 55 L/mol = 5.58 L

Calculate the mass of N2 produced

To find the mass of N2, we need the moles of N2 produced. From the balanced equation, we know that 4 moles of nitroglycerin produce 6 moles of N2. Moles of N2 produced = moles of nitroglycerin × (moles of N2 /moles of nitroglycerin) Mole of N2 produced = 0.0140 mol × (6 mol N2/4 mol nitroglycerin) = 0.0210 mol N2 Now, using the molar mass of N2 (2 × 14.01 g/mol = 28.02 g/mol), we can find the mass of N2 produced: Mass of N2 produced = moles of N2 × molar mass of N2 Mass of N2 produced = 0.0210 mol × 28.02 g/mol = 0.588 g Answer: (a) The number of moles of gas produced is 0.1015 mol. (b) The volume of gas produced is 5.58 L. (c) The mass of N2 produced is 0.588 g.

Key concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. When nitroglycerin detonates, as in the given exercise, it breaks down into different products, including gases like carbon dioxide, nitrogen, oxygen, and water vapor. This transformation is represented by a balanced chemical equation, which ensures the conservation of mass and atoms.

• Each compound on the reactant side must be balanced with the same number of each type of atom on the product side.
• In detonation reactions, energy release can be significant, showing why nitroglycerin is used as an explosive.
• Balanced equations allow us to understand the stoichiometry of the reaction, which helps in calculating quantities of reactants and products.

Understanding these principles helps us make predictions about the amounts of substances involved in the reaction and is essential for calculations in stoichiometry.

Molar Mass

Molar mass is crucial for converting between mass and moles of a substance. It's the mass of one mole of a compound, measured in g/mol, and is pivotal for solving stoichiometry problems.

• To calculate molar mass, sum up the atomic masses of all the atoms in the molecular formula.
• In the exercise, nitroglycerin's molar mass is calculated by adding the masses of carbon, hydrogen, nitrogen, and oxygen atoms in the formula \( \text{C}_3\text{H}_5\text{N}_3\text{O}_9 \).
• This gives us a molar mass of 227.13 g/mol.
• Using this, we can convert the given mass of nitroglycerin into moles by dividing the mass by the molar mass.

Accurate molar mass calculation is vital for determining the correct proportions in chemical equations and understanding the scale of reactions.

Gas Laws

Gas laws explain how gases behave under different conditions. In stoichiometry, understanding these laws helps calculate the volumes of gases produced in reactions. This exercise focuses on the link between moles of gas and the volume they occupy.

• Each mole of gas typically occupies a certain volume—like 55 L in the given conditions—which can be used to convert moles to volume.
• The ideal gas law is a common tool in stoichiometry: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant, and \( T \) is temperature.
• Understanding these relationships helps us predict how the change in one condition (like temperature or pressure) could affect the gas volume.

These principles are fundamental for practical applications, such as calculating the amount of gas produced during an explosive reaction like nitroglycerin's detonation.