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 NaN \(_{3}\) are required to form 10.0 \(\mathrm{g}\) of nitrogen gas? (c) How many grams of 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) 2.25 moles of N2 are produced by the decomposition of 1.50 moles of NaN3. (b) 34.4 g of NaN3 are required to form 10.0 g of nitrogen gas. (c) 1236 g of NaN3 are required to produce 10.0 ft³ of nitrogen gas with a density of 1.25 g/L.

Step-by-step solution

Mole ratio for N2 and NaN3

From the balanced chemical equation, $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g), $$ we can see that every 2 moles of NaN3 will decompose to produce 3 moles of N2 gas. This means that the mole ratio between N2 and NaN3 can be expressed as: $$ \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}}. $$

Determine moles of N2 produced by 1.50 moles of NaN3 (Question a)

Using the mole ratio from Step 1, we can determine the moles of N2 produced by 1.50 moles of NaN3: $$ \text{Moles of N2} = 1.50 \ \text{moles of} \ \mathrm{NaN}_{3} \cdot \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}} = \boxed{2.25 \ \text{moles of} \ \mathrm{N}_{2}}. $$

Molar mass of NaN3 and N2 gases

To answer questions (b) and (c), we need the molar mass of the NaN3 and N2 gases. Molar mass of NaN3: $$ \text{NaN}_{3} = 1 \times (\text{Na}) + 3 \times (\text{N}). $$ From the periodic table, we know the atomic masses of Sodium (Na) and Nitrogen (N) are approximately 23 g/mol and 14 g/mol, respectively. Thus, the molar mass of NaN3 is: $$ \text{Molar Mass of NaN}_{3} = 1 \times (23 \ \text{g/mol}) + 3 \times (14 \ \text{g/mol}) = 23 + 42 = 65 \ \text{g/mol}. $$ Molar mass of N2: $$ \text{N}_{2} = 2 \times (\text{N}). $$ Knowing the atomic mass of Nitrogen (N) is approximately 14 g/mol: $$ \text{Molar Mass of N}_{2} = 2 \times (14 \ \text{g/mol}) = 28 \ \text{g/mol}. $$

Calculate grams of NaN3 needed to form 10.0 g of N2 (Question b)

Using the mole ratio from Step 1 and the molar masses calculated in Step 3, we can determine the grams of NaN3 needed to form 10.0 g of N2: $$ \text{Grams of NaN}_{3} = \frac{10.0 \ \text{g of} \ \mathrm{N}_{2}}{28 \ \text{g/mol}} \cdot \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}} \cdot 65 \ \text{g/mol} \ = \boxed{34.4 \ \text{g of} \ \mathrm{NaN}_{3}}. $$

Calculate grams of NaN3 needed to produce 10.0 ft³ of N2 (Question c)

First, let's convert 10 ft³ of N2 gas into liters. We know that 1 ft³ = 28.317 L: $$ \text{Volume of N}_{2}\text{ in liters} = 10.0 \ \text{ft}^3 \cdot \frac{28.317 \ \text{L}}{1 \ \text{ft}^3} = 283.17 \ \text{L}. $$ Next, we can determine the mass of N2 gas in grams using its density (1.25 g/L): $$ \text{Mass of N}_{2} = 283.17 \ \text{L} \cdot 1.25 \ \text{g/L} = 354 \ \text{g}. $$ Finally, using the mole ratio from Step 1 and the molar masses calculated in Step 3, we can determine the grams of NaN3 needed to produce 354 g of N2: $$ \text{Grams of NaN}_{3} = \frac{354 \ \text{g of} \ \mathrm{N}_{2}}{28 \ \text{g/mol}} \cdot \frac{3 \ \text{moles of} \ \mathrm{N}_{2}}{2 \ \text{moles of} \ \mathrm{NaN}_{3}} \cdot 65 \ \text{g/mol} \ = \boxed{1236 \ \text{g of} \ \mathrm{NaN}_{3}}. $$

Key concepts

Mole Ratio

When dealing with reactions, understanding the mole ratio is key to predicting the amounts of reactants or products. The mole ratio is the relationship between the amounts in moles of any two compounds involved in a chemical reaction. We find mole ratios from a balanced chemical equation. In this case, for the decomposition of sodium azide (\(\mathrm{NaN}_3\)) into sodium and nitrogen gas (\(\mathrm{N}_2\)), the balanced equation is:
\[2 \mathrm{NaN}_3 (s) \rightarrow 2 \mathrm{Na} (s) + 3 \mathrm{N}_2 (g).\]This tells us that 2 moles of sodium azide produce 3 moles of nitrogen gas. This gives a mole ratio for \(\mathrm{N}_2\) and \(\mathrm{NaN}_3\) as \(\frac{3}{2}\), meaning 3 moles of nitrogen gas is generated for every 2 moles of \(\mathrm{NaN}_3\) that undergo decomposition.

• Use the balanced chemical equation to find mole ratios.
• Multiply the given amount by the appropriate mole ratio to find the desired amount.

Molar Mass

Molar mass is the mass of one mole of a substance and is measured in grams per mole (g/mol). It plays a crucial role when converting between the mass of a substance and the amount in moles. To find the molar mass, we sum the atomic masses of all the atoms in a molecule.For sodium azide (\(\mathrm{NaN}_3\)), the molar mass is calculated as follows:

• Molar mass of \(\mathrm{Na}\) = 23 g/mol
• Molar mass of \(\mathrm{N}\) = 14 g/mol
• So, molar mass of \(\mathrm{NaN}_3\) = 23 + (3 \times 14) = 65 g/mol

For nitrogen gas (\(\mathrm{N}_2\)), it is:

• Molar mass of \(\mathrm{N}_2\) = 2 \times 14 = 28 g/mol

Calculating molar mass accurately is essential for determining how much reactant is required to produce a certain amount of product.

Gas Density

Gas density is a measure of a gas's mass per unit volume, usually expressed in grams per liter (g/L). Understanding gas density helps us convert between volume and mass, which is essential in stoichiometry.For example, in the problem, the density of nitrogen gas is given as 1.25 g/L. This information is crucial when calculating how much sodium azide is needed to fill a specific volume of an airbag with nitrogen gas. In scenario (c) of the exercise, if 10.0 ft³ of nitrogen gas is required, the conversion into liters (1 ft³ = 28.317 L) helps in determining the mass:

• Volume of \(\mathrm{N}_2\) in liters = 10 ft³ \times 28.317 L/ft³ = 283.17 L
• Mass of \(\mathrm{N}_2\) = 283.17 L \times 1.25 g/L = 354 g

A firm understanding of gas density allows us to work seamlessly between the different units of volume and mass, crucial for any gas-related stoichiometric calculations.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where one substance breaks down into two or more simpler substances. In the context of this exercise, sodium azide is undergoing decomposition:
\[2 \mathrm{NaN}_3 (s) \rightarrow 2 \mathrm{Na} (s) + 3 \mathrm{N}_2 (g).\]This decomposition is critical for the functionality of airbags, which rapidly inflate by producing nitrogen gas. Decomposition reactions like this often involve energy input in the form of heat, light, or electricity to break chemical bonds.

• Decomposition enables the transformation of a single compound into simpler substances.
• In practical terms, understanding these reactions is important for applications like safety devices, where rapid gas production is needed.
• The balanced chemical equation shows the stoichiometry of the decomposition reaction, essential for calculating reactants and products in such transformations.