Question

Automotive air bags inflate when sodium azide, \(\mathrm{NaN}_{3}\). rapidly decomposes to its component elements: $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{~N}_{2}(g) $$ (a) How many moles of \(\mathrm{N}_{2}\) are produced by the decomposition of \(1.50 \mathrm{~mol}\) of \(\mathrm{Na} \mathrm{N}_{3}\) ? (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

In this stoichiometry problem: (a) 2.25 moles of N2 are produced by the decomposition of 1.50 mol of NaN3. (b) 15.47 grams of NaN3 are required to form 10.0 g of nitrogen gas. (c) 547.95 grams of NaN3 are required to produce 10.0 ft³ of nitrogen gas with a density of 1.25 g/L.

Step-by-step solution

For the given balanced chemical reaction, we have the equation: \( 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s) + 3 \mathrm{N}_{2}(g) \) From this, we can gather that 2 moles of NaN3 produce 3 moles of N2 gas. #Step 2: Solve for moles of N2 produced from 1.50 mol of NaN3 (Part a)#

Using the mole ratio (2:3) of NaN3 and N2, calculate the moles of N2 produced: Moles of N2 = (Moles of NaN3 * 3) / 2 = (1.50 mol * 3) / 2 = 2.25 mol #Step 3: Convert moles of N2 to mass of N2 (Part b)#

The molar mass of N2 is 28 g/mol (given). First, find out the moles of N2 from the given mass (10.0 g): Moles of N2 = 10.0 g / 28 g/mol = 0.357 mol #Step 4: Calculate the mass of NaN3 required to produce a given amount of N2#

Use the mole ratio of NaN3 and N2 to calculate the moles of NaN3 required: Moles of NaN3 = (0.357 mol * 2) / 3 = 0.238 mol Now convert moles of NaN3 to grams using the molar mass of NaN3 (65 g/mol): Mass of NaN3 = 0.238 mol * 65 g/mol = 15.47 g #Step 5: Convert ft³ to L and find the mass of N2 from density (Part c)#

Given density of N2 is 1.25 g/L. First, convert 10.0 ft³ to L: \(1 \mathrm{ft}^{3} = 28.32 \mathrm{L}\) Volume of N2 = 10.0 ft³ * 28.32 L/ft³ = 283.2 L Now, using the density formula, find the mass of N2: Mass of N2 = Density * Volume = 1.25 g/L * 283.2 L = 354 g #Step 6: Calculate the mass of NaN3 required to produce the given volume of N2#

First, find out the moles of N2 from the given mass (354 g): Moles of N2 = 354 g / 28 g/mol = 12.64 mol Use the mole ratio of NaN3 and N2 to calculate the moles of NaN3 required: Moles of NaN3 = (12.64 mol * 2) / 3 = 8.43 mol Now convert moles of NaN3 to grams using the molar mass of NaN3 (65 g/mol): Mass of NaN3 = 8.43 mol * 65 g/mol = 547.95 g

Key concepts

Chemical Reactions

In the world of chemistry, chemical reactions are like intricate dances where substances, known as reactants, transform into new substances called products. Imagine sodium azide, a compound used in airbags, as a performer on stage. When it’s time for the airbag to deploy, the sodium azide swiftly breaks down into elemental sodium and nitrogen gas through a decomposition reaction, a type of chemical reaction.

Understanding these reactions requires a grasp of stoichiometry, which is like the choreography of the reaction. It is the quantitative study of the relative amounts of reactants and products in a chemical reaction. The balanced equation shown in the exercise, 2 NaN₃(s) → 2 Na(s) + 3 N₂(g), tells us that for every two moles of sodium azide decomposed, three moles of nitrogen gas are produced.

This type of quantitative relationship is crucial when making calculations in chemistry, like figuring out how much of a reactant you need to create a certain amount of product, much like the calculations required to determine the correct amount of sodium azide needed to inflate an airbag.

Mole Concept

The mole concept is a fundamental cornerstone in chemistry, akin to counting apples in a basket but for atoms and molecules. One mole is defined as exactly 6.022 x 10²³ particles, which could be atoms, molecules, or ions - this is known as Avogadro's number.

Let's consider part (a) of our exercise. We're asked how many moles of nitrogen gas, N₂, are produced from 1.50 moles of sodium azide, NaN₃. Using the mole concept, we directly apply the ratio from the balanced equation. If 2 moles of NaN₃ yield 3 moles of N₂, then 1.50 moles of NaN₃ will give us 2.25 moles of N₂.

The conversion does not depend on the type of particle but rather on their amount in moles, which serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure and observe.

Molar Mass

The concept of molar mass is like knowing the weight of your suitcase before a flight; it's a crucial number that helps prevent any surprises. In chemistry, the molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol).

Consider our friend sodium azide, NaN₃, once again. Each mole of NaN₃ has a certain mass, which we use to convert between grams and moles of the substance—as showcased when calculating how many grams of NaN₃ are required to produce 10.0 g of nitrogen gas (part b). Using the molar mass of N₂, which is 28 g/mol, we unfold the calculation step by step, eventually finding out that we need 15.47 grams of NaN₃ for our desired quantity of nitrogen.

Understanding molar mass allows us to precisely measure how much of each reactant is needed in a reaction or how much product we can expect. It's the guarantee that reactions go as planned, much like ensuring you've packed the right amount of clothes for the trip – efficient and without any excess.