/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Air bags are activated when a se... [FREE SOLUTION] | 91Ó°ÊÓ

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Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide \(\left(\mathrm{NaN}_{3}\right)\) to decompose explosively according to the following reaction: $$2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$ What mass of \(\mathrm{NaN}_{3}(s)\) must be reacted to inflate an air bag to 70.0 \(\mathrm{L}\) at STP?

Short Answer

Expert verified
To find the mass of sodium azide (NaN3) required to inflate an air bag to 70.0 L at STP, follow these steps: 1. Calculate the moles of N2 gas needed using the Ideal Gas Law: \(n(N_2) = \frac{PV}{RT} = \frac{(1\text{ atm})(70.0\text{ L})}{(0.08206\text{ L atm/mol K})(273.15\text{ K})}\) 2. Determine the moles of NaN3 required using stoichiometry: \(n(\mathrm{NaN}_3) = (n(N_2))\left(\frac{2\text{ mol NaN}_3}{3\text{ mol }N_2}\right)\) 3. Convert moles of NaN3 to mass using molar mass (65.01 g/mol): \(\text{mass}(\mathrm{NaN}_3) = n(\mathrm{NaN}_3) \times \text{molar mass}(\mathrm{NaN}_3)\) After completing these steps, you will find the mass of sodium azide needed to inflate the air bag to 70.0 L at STP.

Step by step solution

01

Determine moles of N2 gas to be generated

First, we must calculate the number of moles of N2 gas that need to be generated to fill the airbag of 70.0 L at STP. We'll use the Ideal Gas Law formula: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearrange the formula to solve for n: n = PV / RT Given that at STP, P = 1 atm and T = 273.15 K, and using the ideal gas constant R = 0.08206 L atm/mol K, we can calculate the moles of N2 gas: n(N2) = (1 atm)(70.0 L) / (0.08206 L atm/mol K)(273.15 K)
02

Calculate moles of NaN3

Now that we have the number of moles of N2 gas required, we must determine the number of moles of NaN3 that need to be reacted according to the balanced chemical equation: $$2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$ Use stoichiometry to calculate the moles of NaN3 needed: n(NaN3) = (n(N2))(2 mol NaN3 / 3 mol N2)
03

Convert moles of NaN3 to mass

Now that we have the moles of NaN3 required, we need to convert it to mass. Use the molar mass of NaN3, which is: Molar mass of NaN3 = 22.99 g/mol (Na) + 14.01 g/mol (N) * 3 = 65.01 g/mol Determine the mass of NaN3 by multiplying the moles of NaN3 by its molar mass: mass(NaN3) = n(NaN3) * molar mass(NaN3) This will give you the mass of sodium azide required to inflate the airbag to 70.0 L at STP.

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Key Concepts

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

Gas Laws
Gas laws describe how gases behave under different conditions of pressure, volume, and temperature. A key formula that helps us understand these behaviors is the Ideal Gas Law, expressed as \( PV = nRT \). Here, \( P \) represents pressure, \( V \) stands for volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant (0.08206 L atm/mol K), and \( T \) is the temperature in Kelvin.

For the airbag problem, we use the Ideal Gas Law to determine how many moles of nitrogen gas are required to fill a 70.0 L airbag at standard temperature and pressure (STP). At STP, the conditions are set at 1 atm pressure and 273.15 K temperature. This simplifies the calculation, since we can easily plug in the values and rearrange the equation to solve for \( n \), the moles of gas.

Understanding these relationships is crucial not just for solving textbook problems, but also for real-world applications where gases need to be contained or managed efficiently and safely.
Chemical Reactions
In chemistry, reactions involve the transformation of substances through breaking and forming chemical bonds. The reaction given in the exercise is: \[2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s) + 3 \mathrm{N}_{2}(g)\]This equation tells us that solid sodium azide (\mathrm{NaN}_{3}) decomposes to produce sodium and nitrogen gas upon activation.

Balancing chemical equations is essential for stoichiometry, as it ensures that atoms are conserved in a reaction. This balance allows us to correlate the quantity of a reactant to the amount of product formed. In our scenario, the decomposition of sodium azide is a critical reaction for the airbag deployment. When triggered, the azide breaks down quickly, releasing nitrogen gas which inflates the bag.

Such chemical equations demonstrate how reactants are transformed and in what proportions, aiding in calculations to identify how much of a substance is needed to achieve the desired product output.
Mole Concept
The concept of the mole is fundamental to chemistry. It provides a bridge between the atomic scale and real-world quantities of chemicals. A mole is the amount of substance that contains as many particles, such as atoms or molecules, as there are atoms in 12 grams of carbon-12. This quantity is known as Avogadro's number, approximately \(6.022 \times 10^{23}\) particles.

In the context of the exercise, the mole concept is used to determine how much \(\mathrm{NaN}_{3}\) is needed to produce a certain number of moles of \(\mathrm{N}_{2}\) gas. From the Ideal Gas Law, we find the required moles of nitrogen, and using the balanced reaction equation, we apply stoichiometry to find out how many moles of sodium azide react to produce it.

Finally, by knowing the molar mass of \(\mathrm{NaN}_{3}\) (65.01 g/mol), we convert moles to grams, aiding in practical applications, such as the precise amount of sodium azide needed for safe airbag deployment. This exemplifies the importance of the mole concept in quantifying and understanding chemical dynamics.

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Most popular questions from this chapter

A chemist weighed out 5.14 g of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50-\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at \(30.0^{\circ} \mathrm{C}\) and 750 . torr. After the reaction to form BaCO_ \(_{3}(s)\) and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was 230 . torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

The oxides of Group 2A metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A 2.85 -g sample containing only MgO and CuO is placed in a \(3.00-\mathrm{L}\) container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\) . After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\) . What is the mass percent of MgO in the mixture? Assume that only the \(\mathrm{MgO}\) reacts with \(\mathrm{CO}_{2}\) .

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 . \mathrm{L} / \mathrm{min}\) at 1.50 \(\mathrm{atm}\) and ambient temperature. Air is added to the chamber at 1.00 \(\mathrm{atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and 79 \(\mathrm{mole}\) percent \(\mathrm{N}_{2},\) calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that 95.0\(\%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\) . The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

A mixture of chromium and zinc weighing 0.362 g was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, 225 mL dry of hydrogen gas was collected at \(27^{\circ} \mathrm{C}\) and \(750 .\) torr. Determine the mass percent of \(\mathrm{Zn}\) in the metal sample [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium (III) chloride and hydrogen gas. \(]\)

A 1;1 mixture by moles of nitrous oxide and oxygen is often used as a sedative in dentistry. If the total pressure of this mixture in a cylinder is 2.50 atm, what is the partial pressure of each gas?

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