Question

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 \(10.0 \mathrm{~g}\) 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

(a) To find the moles of nitrogen gas produced by 1.50 moles of sodium azide, we can set up a proportion using the stoichiometric ratio: $$ \frac{3 \: \text{moles of N}_2}{2 \: \text{moles of NaN}_3} = \frac{x \: \text{moles of N}_2}{1.50 \: \text{moles of NaN}_3} $$ Solving for x yields \(x = 2.25 \: \text{moles of N}_2\). (b) To find the mass of sodium azide required to produce 10.0 grams of nitrogen gas, first find the moles of nitrogen gas: $$ \text{moles of N}_2 = \frac{10.0 \: \text{g} \: \text{N}_2}{28.02 \: \text{g/mol}} \approx 0.357 \: \text{moles of N}_2 $$ Then, use stoichiometry and the molar mass of sodium azide to find the mass in grams: $$ \frac{2 \: \text{moles of NaN}_3}{3 \: \text{moles}} = \frac{y \: \text{moles of NaN}_3}{0.357 \: \text{moles}} $$ Solving for y yields \(y \approx 0.238 \: \text{moles of NaN}_3\), and multiplying by the molar mass gives \(15.44 \: \text{g of NaN}_3\). (c) To find the mass of sodium azide needed to produce 10.0 ft³ of nitrogen gas, first convert the volume to liters: $$ \text{volume in L} = 10.0 \: \text{ft}^3 \times 28.3168 = 283.168 \: \text{L} $$ Next, find the mass of nitrogen gas: $$ \text{mass of N}_2 = 283.168 \: \text{L} \times 1.25 \: \text{g/L} = 353.96 \: \text{g} $$ Finally, follow the approach of step 2 to determine the mass of sodium azide needed. Calculate the moles of nitrogen gas: $$ \text{moles of N}_2 = \frac{353.96 \: \text{g} \: \text{N}_2}{28.02 \: \text{g/mol}} \approx 12.63 \: \text{moles of N}_2 $$ Then, use stoichiometry and the molar mass of sodium azide to find the mass in grams: $$ \frac{2 \: \text{moles of NaN}_3}{3 \: \text{moles}} = \frac{z \: \text{moles of NaN}_3}{12.63 \: \text{moles}} $$ Solving for z yields \(z \approx 8.42 \: \text{moles of NaN}_3\), and multiplying by the molar mass gives \(547.25 \: \text{g of NaN}_3\).

Step-by-step solution

Finding the moles of nitrogen gas in (a)

Given that 2 moles of sodium azide decompose into 3 moles of nitrogen gas, we can use the stoichiometric ratio to calculate how many moles of nitrogen gas are produced by 1.50 moles of sodium azide. Using the given equation: $$ 2 \: \mathrm{NaN}_{3}(s) \longrightarrow 2 \: \mathrm{Na}(s)+3 \: \mathrm{N}_{2}(g) $$ For every 2 moles of sodium azide, 3 moles of nitrogen gas are produced. So for 1.50 moles of sodium azide, we can set up a proportion using the stoichiometric ratio: $$ \frac{3 \: \text{moles of N}_2}{2 \: \text{moles of NaN}_3} = \frac{x \: \text{moles of N}_2}{1.50 \: \text{moles of NaN}_3} $$ Solve for x.

Calculating the mass of sodium azide in (b)

We are given that 10.0 grams of nitrogen gas are produced, and we must find the mass of sodium azide required. To do so, we should first find the moles of nitrogen gas for the given mass, and then use the stoichiometric ratio to calculate the required grams of sodium azide. First, calculate the moles of nitrogen gas produced using its molar mass (\(28.02 \: \text{g/mol}\)): $$ \text{moles of N}_2 = \frac{10.0 \: \text{g} \: \text{N}_2}{28.02 \: \text{g/mol}} $$ Now, use stoichiometry to find the grams of sodium azide required: $$ \frac{2 \: \text{moles of NaN}_3}{3 \: \text{moles of N}_2} = \frac{y \: \text{moles of NaN}_3}{\text{moles of N}_2} $$ Solve for y and multiply by the molar mass of sodium azide (\(65.01 \: \text{g/mol}\)) to find the mass in grams.

Calculating the mass of sodium azide in (c)

We are given that 10.0 ft³ of nitrogen gas with a density of \(1.25 \: \text{g/L}\) is required, and we must find the mass of sodium azide needed. First, we need to convert the volume from ft³ to liters and then use the gas density to find the mass in grams. To convert the volume to liters, use the conversion factor: $$ 1 \: \text{ft}^3 = 28.3168 \: \text{L} $$ Now, find the mass of nitrogen gas using the given density and the volume in liters: $$ \text{mass of N}_2 = \text{volume in L} \times \text{density} $$ Finally, follow the approach of step 2 to determine the mass of sodium azide needed. Calculate the moles of nitrogen gas and then use stoichiometry and the molar mass of sodium azide to find the mass in grams.

Key concepts

Mole Concept

The mole concept is a fundamental cornerstone of chemistry, particularly when dealing with chemical reactions. It offers a method for quantifying substances based on the number of particles, such as atoms, molecules, or ions, rather than their mass. One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\).

When it comes to stoichiometry, the mole concept is essential for converting mass into moles, moles into particles, and for relating the amounts of reactants and products in a balanced chemical equation. For example, in our exercise's chemical decomposition of sodium azide (\(\mathrm{NaN}_3\)), the number of moles of nitrogen gas (\(\mathrm{N}_2\)) produced is directly proportional to the number of moles of sodium azide reacted, following the molar ratio set by the balanced equation.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where a single compound breaks down into two or more simpler substances. The sodium azide used in vehicle airbags decomposing into sodium metal and nitrogen gas is an example of this process. The equation
\(2 \mathrm{NaN}_3(s) \longrightarrow 2 \mathrm{Na}(s) + 3 \mathrm{N}_2(g)\)
represents a decomposition reaction where stoichiometry allows us to determine the mole ratios of reactants and products. In this case, for every two moles of sodium azide that decompose, three moles of nitrogen gas are released. Understanding these ratios is crucial for predicting the amounts of substances produced or required in such reactions.

Molar Mass Calculation

Molar mass, the mass of one mole of a substance, serves as a bridge between the macroscopic world of grams that we can measure and the microscopic world of moles that we use in chemical equations. Calculating molar mass involves summing the atomic masses of all atoms in a molecule. In the provided solution, calculating the molar mass of nitrogen gas (\(\mathrm{N}_2\)) is vital to convert the given 10.0 grams of nitrogen into moles.

For instance, the molar mass of \(\mathrm{N}_2\) can be found by adding the atomic masses of two nitrogen atoms, which comes to about \(28.02 \text{g/mol}\). Knowing this allows us to stratify how many moles of nitrogen gas correspond to a given mass, furthering our ability to solve stoichiometric problems.

Gas Laws and Conversion

The gas laws are principles that predict how gases will behave under different conditions of temperature, pressure, and volume. When the exercise asks about the volume of gas required to inflate an automotive airbag, these laws become significant. To solve for the mass of sodium azide needed to produce a certain volume of gas, we first need to understand the relationship between the volume of a gas and its mass and density. These relationships are part of the Ideal Gas Law.

In this task, we use density and volume to determine the mass of nitrogen gas. Firstly, we convert the volume from cubic feet to liters, acknowledging that one cubic foot equals approximately \(28.3168 \text{L}\). Then, by multiplying the volume by the density of the gas, we calculate the mass. This mass can then be transformed back into moles, and subsequently, the stoichiometric relationships from our balanced equation can be used to determine the mass of reactant required (sodium azide in this case). Such conversions are incredibly useful in predicting the outcomes of reactions involving gases.