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Automotive air bags inflate when sodium azide, \(\mathrm{NaN}_{3}\), rapidly decomposes to its component elements: $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N}_{2}(g) $$ (a) How many moles of \(\mathrm{N}_{2}\) are produced by the decomposition of \(1.50 \mathrm{~mol}\) of \(\mathrm{NaN}_{3}\) ? (b) How many grams of \(\mathrm{NaN}_{3}\) are required to form \(10.0 \mathrm{gg}\) of nitrogen gas? (c) How many grams of \(\mathrm{NaN}_{3}\) are required to produce \(10.0 \mathrm{ft}^{3}\) of nitrogen gas, about the size of an automotive air bag, if the gas has a density of \(1.25 \mathrm{~g} / \mathrm{L}\) ?

Short Answer

Expert verified
(a) 2.25 moles of N2 are produced by the decomposition of 1.50 mol of NaN3. (b) 15.5 grams of NaN3 are required to form 10.0 g of N2 gas. (c) 547.5 grams of NaN3 are required to produce 10.0 ft³ of N2 gas (about the size of an automotive air bag).

Step by step solution

01

Mole-to-Mole Conversion

Using the balanced chemical equation, we can determine the moles of N2 produced by 1.50 mol of NaN3: \(2 \,\text{mol}\, \mathrm{NaN}_{3} \longrightarrow 3 \,\text{mol}\, \mathrm{N}_{2}\) To find the moles of N2 produced, we can use the stoichiometric ratio of moles of NaN3 to moles of N2: Moles of N2 = Moles of NaN3 × (3 moles of N2 / 2 moles of NaN3) = 1.50 mol × (3/2) = \(2.25 \,\text{mol}\, \mathrm{N}_{2}\)
02

Answer (a)

2.25 moles of N2 are produced by the decomposition of 1.50 mol of NaN3. (b)
03

Convert Mass of N2 to Moles

Using the molar mass of N2 (28.02 g/mol), we can convert the mass of N2 to moles: Moles of N2 = (10.0 g) / (28.02 g/mol) = 0.357 mol N2
04

Mole-to-Mole Conversion

Using the balanced chemical equation, we can determine the moles of NaN3 required to produce 0.357 mol of N2: \(2 \,\text{mol}\, \mathrm{NaN}_{3} \longrightarrow 3 \,\text{mol}\, \mathrm{N}_{2}\) Moles of NaN3 = Moles of N2 × (2 moles of NaN3 / 3 moles of N2) = 0.357 mol × (2/3) = 0.238 mol NaN3
05

Convert Moles of NaN3 to Grams

Using the molar mass of NaN3 (65.02 g/mol), we can convert the moles of NaN3 to grams: Mass of NaN3 = Moles of NaN3 × (65.02 g/mol) = 0.238 mol × 65.02 g/mol = 15.5 g NaN3
06

Answer (b)

15.5 grams of NaN3 are required to form 10.0 g of N2 gas. (c)
07

Convert Volume and Density to Mass of N2

First, we must convert the volume of N2 from cubic feet to liters: \(10.0\,\text{ft}^3 × \frac{28.32\, \text{L}}{1\,\text{ft}^3} = 283.2 \, \text{L}\) Then, using the given density of N2 (1.25 g/L), we can find the mass of N2: Mass of N2 = (283.2 L) × (1.25 g/L) = 354.0 g N2
08

Convert Mass of N2 to Moles

Using the molar mass of N2 (28.02 g/mol), we can convert the mass of N2 to moles: Moles of N2 = (354.0 g) / (28.02 g/mol) = 12.64 mol N2
09

Mole-to-Mole Conversion

Using the balanced chemical equation, we can determine the moles of NaN3 required to produce 12.64 mol of N2: \(2 \,\text{mol}\, \mathrm{NaN}_{3} \longrightarrow 3 \,\text{mol}\, \mathrm{N}_{2}\) Moles of NaN3 = Moles of N2 × (2 moles of NaN3 / 3 moles of N2) = 12.64 mol × (2/3) = 8.426 mol NaN3
10

Convert Moles of NaN3 to Grams

Using the molar mass of NaN3 (65.02 g/mol), we can convert the moles of NaN3 to grams: Mass of NaN3 = Moles of NaN3 × (65.02 g/mol) = 8.426 mol × 65.02 g/mol = 547.5 g NaN3
11

Answer (c)

547.5 grams of NaN3 are required to produce 10.0 ft³ of N2 gas (about the size of an automotive air bag).

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Key Concepts

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

Stoichiometry
Stoichiometry is the study of the quantitative relationships between the reactants and products in chemical reactions. It is grounded in the law of conservation of mass that states that in a closed system, matter is neither created nor destroyed. Stoichiometry provides a mathematical means of relating the amounts of reactants to the amounts of products produced.

For example, in the sodium azide decomposition reaction used in automotive airbags, stoichiometry allows us to calculate how many moles of nitrogen gas, \( \mathrm{N}_2 \), are produced from a given quantity of sodium azide, \( \mathrm{NaN}_3 \). The balanced chemical equation shows a ratio of 2 moles of \( \mathrm{NaN}_3 \) producing 3 moles of \( \mathrm{N}_2 \), which sets the foundation for performing calculations to answer specific questions, such as those given in the exercise.

Understanding stoichiometry is essential for predicting the outcomes of chemical reactions and for scaling the production of desired substances in various industrial processes.
Molar Mass Conversion
Molar mass conversion is a pivotal aspect of chemistry that involves converting between mass and moles of a substance using its molar mass. The molar mass is defined as the mass of one mole of a given substance and is measured in grams per mole \( \text{g/mol} \).

As shown in the exercise, to find the mass of sodium azide required to produce a specific amount of nitrogen gas, you need to know the molar masses of both substances: \( \mathrm{NaN}_3 \) and \( \mathrm{N}_2 \). With molar masses, scientists can relate the mass of a substance in grams to the amount in moles, thereby quantifying substances in chemical reactions.

For practitioners, mastering molar mass conversions ensures accurate measurements and calculations when preparing chemical solutions, conducting experiments, or analyzing reaction yields.
Chemical Reaction Balancing
Chemical reaction balancing is the process of ensuring that the number of atoms of each element is conserved across the reactants and products in a chemical equation. In accordance with the law of conservation of mass, a balanced chemical equation must have the same number of each type of atom on both sides of the reaction.

In the sodium azide decomposition reaction, the balanced equation \( 2 \mathrm{NaN}_3(s) \longrightarrow 2 \mathrm{Na}(s) + 3 \mathrm{N}_2(g) \) shows that two sodium azide units produce two sodium atoms and three nitrogen gas molecules. Balancing chemical equations is fundamental for predicting the amounts of substances produced or consumed in a reaction, vital for both academic problem-solving and real-world applications like the manufacturing of airbags.
Gas Density Conversion
Gas density conversion involves the use of a substance's density to relate its volume to its mass. Density is defined as the mass per unit volume of a substance and is expressed in \( \text{g/L} \) for gases. In practical scenarios, such as defining the size of an airbag in terms of the gas it contains, chemists convert the volume of the gas to mass using the known density.

In our example, the density of nitrogen gas allows us to determine the mass of \( \mathrm{N}_2 \) required to fill a 10.0 cubic foot airbag. Converting from cubic feet to liters and then using the density to find the mass enables further stoichiometric calculations. Understanding gas density conversion is critical in applications spanning from industrial gas storage and transportation to environmental monitoring of pollutants.

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Most popular questions from this chapter

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