Problem 70 - Chemistry The Central Science

Question

Detonation of nitroglycerin proceeds as follows: $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}_{3} \mathrm{O}_{9}(l) \longrightarrow \\ \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad12 \mathrm{CO}_{2}(g)+6 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) If a sample containing 2.00 \(\mathrm{mL}\) of nitroglycerin (density \(=\) 1.592 \(\mathrm{g} / \mathrm{mL}\) ) is detonated, how many moles of gas are produced? (b) If each mole of gas occupies 55 Lunder 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

(a) From the detonation of 2.00 mL of nitroglycerin, 0.406 moles of gas are produced. (b) The volume of gas produced under the explosion conditions is 22.33 L. (c) In the detonation, 0.5893 grams of nitrogen gas are produced.

Step 1: Calculate the mass of nitroglycerin

Given the volume of nitroglycerin is 2.00 mL and the density is 1.592 g/mL. Calculate the mass of nitroglycerin by using the following formula: Mass = Volume x Density Substitute the given values: \(Mass = 2.00 \, \mathrm{mL} \times 1.592 \frac{\mathrm{g}}{\mathrm{mL}}\) Calculate the mass of nitroglycerin: \(Mass = 3.184 \, \mathrm{g}\) The mass of nitroglycerin used is 3.184 grams.

Step 2: Convert grams to moles

To calculate the moles of nitroglycerin, find the molar mass (MM) of nitroglycerin, C3H5N3O9: MM = 3(M_C) + 5(M_H) + 3(M_N) + 9(M_O) Substitute the molar masses for each element: MM = 3(12.01 g/mol) + 5(1.008 g/mol) + 3(14.01 g/mol) + 9(16.00 g/mol) Compute the molar mass: MM = 227.09 g/mol Now, find the moles of nitroglycerin, using the formula: Moles = Mass / MM Substitute the mass and molar mass values: \(Moles = \frac{3.184 \, \mathrm{g}}{227.09 \, \mathrm{g/mol}}\) Calculate the moles of nitroglycerin: \(Moles = 0.01402 \, \mathrm{mol}\) There are 0.01402 moles of nitroglycerin in the sample.

Step 3: Calculate moles of gas produced

Using the stoichiometry of the balanced chemical equation, calculate the moles of gas produced: 1 moles of nitroglycerin produce 12 moles of CO2 + 6 moles of N2 + 1 moles of O2 + 10 moles of H2O = 29 moles of gas Therefore, the moles of gas produced are: Moles of gas = 0.01402 mol nitroglycerin x 29 moles of gas / 1 mol nitroglycerin Calculate the moles of gas produced: Moles of gas = 0.406 \, \mathrm{mol}\ A total of 0.406 moles of gas are produced in the detonation.

Step 4: Calculate the volume of gas produced

Given that each mole of gas occupies 55 L under the explosion conditions, we can use the relationship between moles and volume: Volume of gas = Moles of gas x 55 L/mol Substitute the value for moles of gas: \(Volume \, of \, gas = 0.406 \, \mathrm{mol} \times 55 \, \frac{\mathrm{L}}{\mathrm{mol}}\) Calculate the volume of gas produced: Volume of gas = 22.33 L The volume of gas produced upon detonation is 22.33 liters.

Step 5: Calculate grams of nitrogen gas produced

To calculate the grams of N2 produced, first find the moles of N2: Given the stoichiometry of the balanced chemical reaction: 4 moles of nitroglycerin produce 6 moles of N2 Calculate the moles of N2 produced: Moles of N2 = 0.01402 mol nitroglycerin x (6 moles of N2 / 4 moles of nitroglycerin) Moles of N2 = 0.02103 mol Now, convert moles of N2 to grams using the molar mass of N2 (28.02 g/mol): Mass of N2 = Moles of N2 x Molar mass of N2 Substitute the calculated moles and molar mass: \(Mass \, of \, N_{2} = 0.02103 \, \mathrm{mol} \times 28.02 \, \frac{\mathrm{g}}{\mathrm{mol}}\) Calculate the mass of nitrogen gas produced: Mass of N2 = 0.5893 g A total of 0.5893 grams of nitrogen gas are produced in the detonation.