Problem 67 - Chemistry The Central Science In Si Units

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 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 \(1.25 \mathrm{~g} / \mathrm{L}\) ?

Step-by-Step Solution

Answer

(a) 2.25 mol of \(\mathrm{N}_{2}\), (b) 15.5 g of \(\mathrm{NaN}_{3}\), (c) 547.65 g of \(\mathrm{NaN}_{3}\).

Step 1: Understand the Reaction and Given Information

The balanced reaction is \(2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)\). For every 2 moles of \(\mathrm{NaN}_{3}\) decomposed, 3 moles of \(\mathrm{N}_{2}\) are produced. We'll find how much \(\mathrm{N}_{2}\) is produced or how much \(\mathrm{NaN}_{3}\) is needed based on this stoichiometry.

Step 2: Calculate Moles of \(\mathrm{N}_{2}\) for a)

From the reaction stoichiometry, 2 moles of \(\mathrm{NaN}_{3}\) produce 3 moles of \(\mathrm{N}_{2}\). Use the ratio to find moles of \(\mathrm{N}_{2}\) from 1.50 moles of \(\mathrm{NaN}_{3}\):\[ \text{Moles of } \mathrm{N}_{2} = 1.50 \, \text{mol} \, \mathrm{NaN}_{3} \times \frac{3 \, \text{mol} \mathrm{N}_{2}}{2 \, \text{mol} \mathrm{NaN}_{3}} = 2.25 \, \text{mol} \, \mathrm{N}_{2}. \]

Step 3: Determine Grams of \(\mathrm{NaN}_{3}\) for b)

First, convert \(10.0 \, \text{g}\) of \(\mathrm{N}_{2}\) to moles using molar mass \(\mathrm{(N}_{2}=28.02 \, \text{g/mol)}\): \[10.0 \, \text{g} \times \frac{1 \, \text{mol}}{28.02 \, \text{g}} = 0.357 \, \text{mol} \, \mathrm{N}_{2}\] Using the reaction stoichiometry, find moles of \(\mathrm{NaN}_{3}\) needed: \[0.357 \, \text{mol} \, \mathrm{N}_{2} \times \frac{2 \, \text{mol} \mathrm{NaN}_{3}}{3 \, \text{mol} \mathrm{N}_{2}} = 0.238 \, \text{mol} \, \mathrm{NaN}_{3}. \] Convert to grams using molar mass of \(\mathrm{NaN}_{3}=65.01 \, \text{g/mol}:\) \[0.238 \, \text{mol} \times 65.01 \, \text{g/mol} = 15.5 \, \text{g}.\]

Step 4: Find Grams of \(\mathrm{NaN}_{3}\) for c)

First, convert \(10.0 \, \text{ft}^{3}\) to liters: \(10.0 \, \text{ft}^{3} = 283.17 \, \text{L}.\) Calculate the mass of \(\mathrm{N}_{2}\) using the density: \[283.17 \, \text{L} \times 1.25 \, \text{g/L} = 353.96 \, \text{g} \, \mathrm{N}_{2}. \] Convert grams to moles: \[353.96 \, \text{g} \times \frac{1 \, \text{mol}}{28.02 \, \text{g}} = 12.63 \, \text{mol} \, \mathrm{N}_{2}.\] Find moles of \(\mathrm{NaN}_{3}\) and convert to grams: \[12.63 \, \text{mol} \, \mathrm{N}_{2} \times \frac{2 \, \text{mol} \mathrm{NaN}_{3}}{3 \, \text{mol} \mathrm{N}_{2}} = 8.42 \, \text{mol} \, \mathrm{NaN}_{3}.\] \[8.42 \, \text{mol} \times 65.01 \, \text{g/mol} = 547.65 \, \text{g}.\]

Key Concepts

Decomposition Reaction

A decomposition reaction occurs when a single compound splits into two or more simpler substances. In chemical terms, it's like taking apart a complicated puzzle to see the smaller pieces that make it whole.

During the decomposition of sodium azide (\(\mathrm{NaN}_3\)), used in automotive airbags, the compound breaks down into sodium (\(\mathrm{Na}\)) and nitrogen gas (\(\mathrm{N}_2\)). The balanced chemical equation is:

• \(2 \mathrm{NaN}_3(s) \rightarrow 2 \mathrm{Na}(s) + 3 \mathrm{N}_2(g)\)

This means 2 moles of sodium azide produce 3 moles of nitrogen gas.

Understanding this reaction is the first crucial step when solving stoichiometric problems. It tells us how much of each substance is involved and produced during the reaction. When sodium azide decomposes, a reaction essential for inflating airbags, it provides a swift release of gas ensuring rapid inflation. By knowing the stoichiometry, i.e., the quantitative relationship between the reactants and products, we can predict how much nitrogen gas is formed from a given amount of sodium azide.

Mole-to-Mole Conversion

Mole-to-mole conversions allow us to determine quantities of products and reactants in a chemical reaction, based on their stoichiometric coefficients.

Let's look at the reaction for sodium azide decomposition. The stoichiometric coefficients from the balanced equation \(2 \mathrm{NaN}_3 \rightarrow 2 \mathrm{Na} + 3 \mathrm{N}_2\) show that to produce 3 moles of nitrogen gas (\(\mathrm{N}_2\)), we need 2 moles of sodium azide (\(\mathrm{NaN}_3\)).

If you start with 1.50 moles of \(\mathrm{NaN}_3\), how much \(\mathrm{N}_2\) will you get? The answer lies in the conversion ratio derived from the balanced equation:

• Conversion ratio: \(\frac{3 \text{ mol } \mathrm{N}_2}{2 \text{ mol } \mathrm{NaN}_3}\)

By multiplying the moles of \(\mathrm{NaN}_3\) by this ratio:

• \(1.50 \text{ mol } \mathrm{NaN}_3 \times \frac{3 \text{ mol } \mathrm{N}_2}{2 \text{ mol } \mathrm{NaN}_3} = 2.25 \text{ mol } \mathrm{N}_2\)

This type of conversion is foundational in chemistry, useful in both laboratory settings and practical applications, such as ensuring airbags deploy correctly by producing an adequate amount of gas.

Gas Density

Gas density relates the mass of a gas to its volume and is a vital concept in determining how much of a solid reactant is necessary to produce a specific gas volume.

Density is typically expressed as grams per liter (g/L), and it shows how much gas is contained within a certain volume. For nitrogen gas (\(\mathrm{N}_2\)), if we have a known density of 1.25 g/L, we can assess the mass for any volume.

In practical scenarios, such as determining how much sodium azide is needed to inflate a car airbag to occupy 10.0 ft³, start by converting the volume to liters. Knowing:

• \(10.0 \text{ ft}^3 = 283.17 \text{ L}\)

Then, apply the density to find the total mass:

• \(283.17 \text{ L} \times 1.25 \text{ g/L} = 353.96 \text{ g } \mathrm{N}_2\)

Convert this mass to moles using the molar mass of \(\mathrm{N}_2\), which is 28.02 g/mol, and proceed with stoichiometry to find out how much sodium azide is required. This approach highlights the importance of understanding gas properties and stoichiometry to ensure car safety features work as intended.