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Sodium azide, the explosive compound in automobile air bags, decomposes according to the following equation: $$2 \mathrm{NaN}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_{2}(\mathrm{g})$$ What mass of sodium azide is required to provide the nitrogen needed to inflate a \(75.0-\mathrm{L}\) bag to a pressure of 1.3 atm at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
171.63 grams of sodium azide are needed.

Step by step solution

01

Calculate moles of nitrogen gas required

First, determine the number of moles of nitrogen gas, \( N_2 \), needed to fill the airbag. Use the Ideal Gas Law: \[PV = nRT\]where \( P \) is the pressure (1.3 atm), \( V \) is the volume (75.0 L), \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is the temperature in Kelvin (25°C + 273 = 298 K). Solve for \( n \), the number of moles:\[n = \frac{PV}{RT} = \frac{1.3 \times 75.0}{0.0821 \times 298} = 3.96 \text{ moles of } N_2\]
02

Relate moles of nitrogen gas to moles of sodium azide

The decomposition reaction shows that 3 moles of \( N_2 \) are produced from 2 moles of \( \text{NaN}_3 \). Thus, calculate the moles of \( \text{NaN}_3 \) needed:\[\frac{2\, \text{moles of } \text{NaN}_3}{3\, \text{moles of } N_2} = \frac{x}{3.96}\implies x = \frac{2 \times 3.96}{3} = 2.64 \text{ moles of } \text{NaN}_3\]
03

Calculate the mass of sodium azide

Finally, find the mass of sodium azide required. The molar mass of \( \text{NaN}_3 \) is 65.01 g/mol. Therefore:\[m = n \times \text{molar mass} = 2.64 \times 65.01 = 171.63 \text{ grams}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Azide
Sodium azide (NaN₃) is a crucial chemical compound used in automotive airbags. It is known for its explosive properties. When triggered, sodium azide decomposes rapidly, producing a large volume of nitrogen gas (N₂). This swift reaction inflates the airbag, ensuring safety during collisions.
Sodium azide's role as an airbag inflator is due to its ability to decompose into sodium (Na) and nitrogen gas, as highlighted in the equation:
  • 2 NaN₃(s) → 2 Na(s) + 3 N₂(g)
This reaction is valued for its efficiency and speed, crucial for immediate airbag deployment.
Understanding sodium azide's decomposition is essential when examining its application in safety devices. Its properties and reactions underscore its significance in modern automotive design.
Mole Calculations
Mole calculations are a fundamental part of chemistry used to determine the amount of a substance. They are essential in understanding chemical reactions and relationships between different compounds.
The mole concept revolves around Avogadro's number, 6.022 x 10²³, representing the number of atoms or molecules in one mole of a substance. In the context of the sodium azide reaction, we calculate the moles of substances involved to ensure precise quantification.
In our example, the moles of nitrogen gas ( N₂) required to fill an airbag are calculated using the Ideal Gas Law (PV = nRT), where we determine the moles needed to produce the right amount of nitrogen gas, ensuring the system's efficiency. Accurate mole calculations like these are crucial when translating theoretical chemistry knowledge into practical applications, such as safety equipment design.
Chemical Decomposition
Chemical decomposition is a type of chemical reaction where a single compound breaks down into two or more simpler substances. It plays a significant role in various industrial and laboratory processes.
In the decomposition of sodium azide (NaN₃), the compound breaks down into sodium metal (Na) and nitrogen gas (N₂). This is a classic example of decomposition, with the equation being:
  • 2 NaN₃(s) → 2 Na(s) + 3 N₂(g)
This reaction is exothermic, meaning it releases energy, thus facilitating rapid decomposition and gas production. Understanding chemical decomposition is key when analyzing chemical processes that require quick or large-scale gas production.
In practical terms, such decomposition reactions are harnessed in airbags to ensure immediate inflation, a critical aspect of vehicle safety mechanisms.
Gas Laws
Gas laws are mathematical relationships and principles that describe the behavior of gases under various conditions. They are crucial for predicting how gases act in different environments.
The Ideal Gas Law is one of the most important gas laws used to solve problems involving gases. It combines several previous gas laws into one equation: PV = nRT, where:
  • P is the pressure of the gas (in atm)
  • V is the volume (in liters)
  • n is the number of moles
  • R is the universal gas constant (0.0821 L·atm/mol·K)
  • T is the temperature (in Kelvin)
In the exercise, the Ideal Gas Law was used to calculate the moles of nitrogen gas ( N₂) required to inflate an airbag. Each variable must be matched with its corresponding unit for accuracy.
Knowledge of gas laws enables predictions about how changes in pressure, temperature, or volume can affect gas behavior, which is particularly useful in designing safety features like airbags.

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Most popular questions from this chapter

Nitrogen trifluoride is prepared by the reaction of ammonia and fluorine. $$ 4 \mathrm{NH}_{3}(\mathrm{g})+3 \mathrm{F}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{NH}_{4} \mathrm{F}(\mathrm{s})+\mathrm{NF}_{3}(\mathrm{g}) $$ If you mix \(\mathrm{NH}_{3}\) with \(\mathrm{F}_{2}\) in the correct stoichiometric ratio, and if the total pressure of the mixture is \(120 \mathrm{mm} \mathrm{Hg}\), what are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{F}_{2} ?\) When the reactants have been completely consumed, what is the total pressure in the flask? (Assume \(T\) is constant.)

The density of air \(20 \mathrm{km}\) above Earth's surface is \(92 \mathrm{g} / \mathrm{m}^{3} .\) The pressure of the atmosphere is \(42 \mathrm{mm} \mathrm{Hg}\) and the temperature is \(-63^{\circ} \mathrm{C}\) (a) What is the average molar mass of the atmosphere at this altitude? (b) If the atmosphere at this altitude consists of only \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2},\) what is the mole fraction of each gas?

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg}\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pressure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

If you place \(2.25 \mathrm{g}\) of solid silicon in a \(6.56-\mathrm{L}\). flask that contains \(\mathrm{CH}_{3} \mathrm{Cl}\) with a pressure of \(585 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\) what mass of dimethyldichlorosilane, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) can be formed? $$\mathrm{Si}(\mathrm{s})+2 \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g}) \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})$$ What pressure of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}(\mathrm{g})\) would you expect in this same flask at \(95^{\circ} \mathrm{C}\) on completion of the reaction? (Dimethyldichlorosilane is one starting material used to make silicones, polymeric substances used as lubricants, antistick agents, and water-proofing caulk.)

If equal masses of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) are placed in separate containers of equal volume at the same temperature, which of the following statements is true? If false, explain why it is false. (a) The pressure in the flask containing \(\mathrm{N}_{2}\) is greater than that in the flask containing \(\mathbf{O}_{2}\) (b) There are more molecules in the flask containing \(\mathrm{O}_{2}\) than in the flask containing \(\mathrm{N}_{2}\)

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