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Chapter 3: Problem 70

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

Expert verified
(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

01

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
02

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
03

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
04

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.

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