Chemical explosions are characterized by the instantaneous release of large quantities of hot gases, which set up a shock wave of enormous pressure (up to 700,000 atm) and velocity (up to 20,000 mi/h). For example, explosion of nitroglycerin (C 3 H 5 N 3 O 9 ) releases four gases, A, B, C, and D: n C 3 H 5 N 3 O 9 (l) a A(g) + b B(g) + c C(g) + d D(g) Assume that the explosion of 1 mol (227 g) of nitroglycerin releases gases with a temperature of 1950 °C and a volume of 1323 L at 1.00 atm pressure. (e) Write a balanced equation for the explosion of nitroglycerin.

Hello everyone. So in this video we're trying to balance out the reaction for the detonation of TNT, which is tri nitro valuing. So of course then our story material becomes TNT, which is composed of C seven H 5 and 306. And what it detonates too is N two H two Carbon monoxide and carbon. Alright, so what I like to do is draw a line first to kind of separate the products from our starting reagents and then write out how much of each material or adam I have. So we have carbons, hydrogen, nitrogen and oxygen of course. Same thing on the product site. Alright, so for the left side we have seven carbons. We have five high regions, three nitrogen and six oxygen's for the right side, I have three carbons to hydrogen to nitrogen is and one oxygen. So the first one that stands out to me will be the bounds of nitrogen. There's no one number that we can add to one side to give us the other side's number of atoms. So we can do then is on the left, on the right on the left side. We multiply this by two and then for the right side we can multiply this by three. So now we do this, we can see that the nitrogen will give us six nitrogen per each side and then not cuddling for the rest of the items on the left side. So we have a coefficient of two. We're multiplying everything by two carbons will not become 14 for the hydrogen is we now have 10 and the options now we have 12. So what I want to go ahead and balance out next is going to be our oxygen's. So we only have one source of oxygen on the right side, that's our carbon monoxide. So we'll go ahead and add the coefficient of 12. So we now get 12 oxygen's and then for our carbons, we now have 13. So that naturally we want to go ahead and manipulate our carbons. So the only single one that we can go ahead and make this 13 to 14 will be this carbon here. So we add a coefficient of two That gives us Carbons. Lastly, we need to balance out our ox hydrogen. So we have 10 and left and two on the right. So we can multiply the right by five to give us 10 hydrogen is on the right side. So now for our final reaction, our overall reaction and balanced equation, most importantly, is the two C seven H five N 306, which is of course just two moles of TNT. And when it detonates, it will give us three moles of N 25 moles of H two, 12 moles of CO and two moles of carbon. And that is going to be my final answer for this problem. Thank you all so much for watching

Ch.10 - Gases: Their Properties & Behavior

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McMurry 8th Edition

Ch.10 - Gases: Their Properties & Behavior

Problem 149e

Chapter 10, Problem 149e

Chemical explosions are characterized by the instantaneous release of large quantities of hot gases, which set up a shock wave of enormous pressure (up to 700,000 atm) and velocity (up to 20,000 mi/h). For example, explosion of nitroglycerin (C₃H₅N₃O₉) releases four gases, A, B, C, and D:
n C₃H₅N₃O₉(l) a A(g) + b B(g) + c C(g) + d D(g)
Assume that the explosion of 1 mol (227 g) of nitroglycerin releases gases with a temperature of 1950 °C and a volume of 1323 L at 1.00 atm pressure.
(e) Write a balanced equation for the explosion of nitroglycerin.

Verified step by step guidance

Identify the chemical formula for nitroglycerin: C_3H_5N_3O_9.

Determine the products of the explosion. Typically, nitroglycerin decomposes into gases such as CO_2, H_2O, N_2, and O_2.

Write the unbalanced chemical equation: C_3H_5N_3O_9(l) -> a CO_2(g) + b H_2O(g) + c N_2(g) + d O_2(g).

Balance the equation by ensuring the number of atoms of each element is equal on both sides. Start with carbon, then hydrogen, nitrogen, and finally oxygen.

Verify the balanced equation by counting the atoms of each element on both sides to ensure they match.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a reaction based on balanced chemical equations. Understanding stoichiometry is essential for writing balanced equations, as it ensures that the law of conservation of mass is upheld.

Gas Laws

Gas laws describe the behavior of gases in relation to pressure, volume, and temperature. Key laws include Boyle's Law, Charles's Law, and the Ideal Gas Law, which can be used to relate the conditions of gases produced in a reaction. In the context of explosions, these laws help in understanding how the gases behave under extreme conditions, such as high temperatures and pressures.

Thermochemistry

Thermochemistry is the study of the heat energy associated with chemical reactions and phase changes. It involves understanding how energy is absorbed or released during a reaction, which is crucial for explosive reactions like that of nitroglycerin. Knowledge of thermochemistry helps in predicting the temperature and energy changes that occur during the explosion, which is vital for safety and control in chemical processes.

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Textbook Question

The Rankine temperature scale used in engineering is to the Fahrenheit scale as the Kelvin scale is to the Celsius scale. That is, 1 Rankine degree is the same size as 1 Fahrenheit degree, and 0 °R = absolute zero. (b) What is the value of the gas constant R on the Rankine scale in 1L ~ atm2>1°R ~ mol2? (c) Use the van der Waals equation to determine the pressure inside a 400.0-mL vessel that contains 2.50 mol of CH4 at a temperature of 525 °R. For CH4, a = 2.253 1L2 ~ atm2>mol2 and b = 0.04278 L>mol.

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Textbook Question

Chemical explosions are characterized by the instantaneous release of large quantities of hot gases, which set up a shock wave of enormous pressure (up to 700,000 atm) and velocity (up to 20,000 mi/h). For example, explosion of nitroglycerin (C₃H₅N₃O₉) releases four gases, A, B, C, and D:

n C₃H₅N₃O₉(l) a A(g) + b B(g) + c C(g) + d D(g)

Assume that the explosion of 1 mol (227 g) of nitroglycerin releases gases with a temperature of 1950 °C and a volume of 1323 L at 1.00 atm pressure.

(d) When gases C and D were passed through a hot tube of powdered copper, gas C reacted to form CuO. The remaining gas, D, was collected in a third 500.0-mL flask and found to have a mass of 0.168 g and a pressure of 223 mm Hg at 25 °C. How many moles each of C and D were present, and what are their likely identities?

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Textbook Question

Combustion analysis of 0.1500 g of methyl tert-butyl ether, an octane booster used in gasoline, gave 0.3744 g of CO2 and 0.1838 g of H2O. When a flask having a volume of 1.00 L was evacuated and then filled with methyl tertbutyl ether vapor at a pressure of 100.0 kPa and a temperature of 54.8 °C, the mass of the flask increased by 3.233 g.(b) What is the molecular weight and molecular formula of methyl tert-butyl ether?

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Textbook Question

Combustion analysis of 0.1500 g of methyl tert-butyl ether, an octane booster used in gasoline, gave 0.3744 g of CO₂ and 0.1838 g of H₂O. When a flask having a volume of 1.00 L was evacuated and then filled with methyl tertbutyl ether vapor at a pressure of 100.0 kPa and a temperature of 54.8 °C, the mass of the flask increased by 3.233 g. (a) What is the empirical formula of methyl tert-butyl ether?

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Textbook Question

Combustion analysis of 0.1500 g of methyl tert-butyl ether, an octane booster used in gasoline, gave 0.3744 g of CO₂ and 0.1838 g of H₂O. When a flask having a volume of 1.00 L was evacuated and then filled with methyl tertbutyl ether vapor at a pressure of 100.0 kPa and a temperature of 54.8 °C, the mass of the flask increased by 3.233 g. (c) Write a balanced equation for the combustion reaction.

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