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Chapter 5: Problem 65

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 \(\operatorname{NaN}_{3}(s)\) must be reacted to inflate an air bag to \(70.0 \mathrm{~L}\) at STP?

Short Answer

Expert verified
Approximately 136.54 g of sodium azide (NaN3) is required to inflate the airbag to 70.0 L at STP.

Step by step solution

01

Define the given parameters and Ideal Gas Law

: Given volume of airbag, V = 70.0 L At STP, temperature T = 273.15 K, and pressure P = 1 atm. The Ideal Gas Law equation is: PV = nRT Where P is pressure, V is volume, n is the number of moles of the gas, R is the Ideal Gas Law constant (0.0821 L atm / mol K), and T is the temperature in Kelvin.
02

Calculate the number of moles of N2 gas

: Using the Ideal Gas Law equation, we can calculate the number of moles (n) of N2 gas required to inflate the airbag. 1 atm * 70.0 L = n * 0.0821 L atm / mol K * 273.15 K Now, solve for n: n = (1 atm * 70.0 L) / (0.0821 L atm / mol K * 273.15 K) n ≈ 3.15 moles of N2(g)
03

Convert moles of N2 gas to moles of NaN3

: Using the balanced chemical equation, we know that 3 moles of N2(g) are produced from 2 moles of NaN3(s). So we can set up a proportion to convert moles of N2(g) to moles of NaN3(s). (3.15 moles of N2) / X moles of NaN3 = 3 moles of N2 / 2 moles of NaN3 Solve for X: X moles of NaN3 = (3.15 moles of N2 * 2 moles of NaN3) / 3 moles of N2 X ≈ 2.10 moles of NaN3(s)
04

Calculate the mass of NaN3 required

: Finally, we'll calculate the mass of NaN3 required by multiplying the moles of NaN3 by its molar mass. Molar mass of NaN3 = (22.99 g/mol for Na) + (3 * 14.01 g/mol for N) = 65.02 g/mol Mass of NaN3 = (2.10 moles of NaN3) * (65.02 g/mol) Mass of NaN3 ≈ 136.54 g Therefore, approximately 136.54 g of sodium azide (NaN3) is required to inflate the airbag to 70.0 L at STP.

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Most popular questions from this chapter

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose \(240 . \mathrm{mL}\) of hydrogen gas is collected at \(30 .{ }^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

Draw a qualitative graph to show how the first property varies with the second in each of the following (assume 1 mole of an ideal gas and \(T\) in kelvins). a. \(P V\) versus \(V\) with constant \(T\) b. \(P\) versus \(T\) with constant \(\underline{V}\) c. \(T\) versus \(V\) with constant \(P\) d. \(P\) versus \(V\) with constant \(T\) e. \(P\) versus \(1 / V\) with constant \(T\) f. \(P V / T\) versus \(P\)

An \(11.2-\mathrm{L}\) sample of gas is determined to contain \(0.50 \mathrm{~mol} \mathrm{~N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: Ammonia gas at \(223^{\circ} \mathrm{C}\) and 90 . atm flows into a reactor at a rate of \(500 . \mathrm{L} / \mathrm{min}\). Carbon dioxide at \(223^{\circ} \mathrm{C}\) and 45 atm flows into the reactor at a rate of \(600 . \mathrm{L} / \mathrm{min}\). What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Atmospheric scientists often use mixing ratios to express the concentrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): ppmv of \(X=\frac{\text { vol of } X \text { at STP }}{\text { total vol of air at STP }} \times 10^{6}\) On a certain November day the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached \(3.0 \times 10^{2}\) ppmv. The atmospheric pressure at that time was 628 torr and the temperature was \(0^{\circ} \mathrm{C}\). a. What was the partial pressure of \(\mathrm{CO} ?\) b. What was the concentration of \(\mathrm{CO}\) in molecules per cubic meter? c. What was the concentration of \(\mathrm{CO}\) in molecules per cubic centimeter?
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