Question

One of the reactions used to inflate automobile air bags involves sodium azide \(\left(\mathrm{NaN}_{3}\right) : 2 \mathrm{NaN}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_{2}(\mathrm{g})\) mass of \(\mathrm{N}_{2}\) produced from the decomposition of NaN \(_{3}\) shown at right.

Short answer

Given a mass (X g) of sodium azide (NaN3), the mass of nitrogen gas (N2) produced can be calculated using the balanced chemical equation and stoichiometry: Mass of N2 produced = \(\frac{1.5 \times X \: \text g \times 28.02 \: \text g/mol}{65.01 \: \text g/mol}\)

Step-by-step solution

Write down the balanced chemical equation

The given balanced chemical equation is: \(2 \: \mathrm{NaN}_{3}(\mathrm{s}) \rightarrow 2 \: \mathrm{Na}(\mathrm{s})+3 \: \mathrm{N}_{2}(\mathrm{g})\)

Determine the molar mass of NaN3 and N2

Using the periodic table, we can find the molar mass for each chemical species in the reaction. For NaN3: Molar mass = 1 mol Na(22.99 g/mol) + 3 mol N(14.01 g/mol) = 65.01 g/mol For N2: Molar mass = 2 mol N(14.01 g/mol) = 28.02 g/mol

Convert mass of NaN3 to moles of NaN3

To convert the mass of NaN3 to moles, we can use the molar mass calculated in Step 2. Let's assume that we have been given X grams of NaN3. Moles of NaN3 = (X g) / (65.01 g/mol)

Determine moles of N2 produced by decomposition of NaN3

Using the balanced chemical equation in Step 1, we can determine the stoichiometric relationship between NaN3 and N2: 2 moles of NaN3 produce 3 moles of N2 So, 1 mole of NaN3 will produce 1.5(=3 / 2) moles of N2. Therefore, Moles of N2 produced = 1.5 × Moles of NaN3 = 1.5 × (X g / 65.01 g/mol)

Convert moles of N2 to mass of N2

Now we will use the molar mass of N2 calculated in Step 2 to convert the moles of N2 to the mass of N2 produced. Mass of N2 produced = Moles of N2 × Molar Mass of N2 = 1.5 × (X g / 65.01 g/mol) × 28.02 g/mol

Simplify the expression for the mass of N2 produced

We will simplify the expression to find the mass of N2 produced for a given mass (X g) of NaN3. Mass of N2 produced = (1.5 × X g × 28.02 g/mol )/(65.01 g/mol) Now we have an expression to calculate the mass of nitrogen gas (N2) produced from a given mass of sodium azide (NaN3).

Key concepts

Balanced Chemical Equations

Understanding balanced chemical equations is crucial for comprehending the intricacies of chemical reactions. In essence, a balanced chemical equation ensures that the number of atoms for each element is the same on both the reactants' side and the products' side of the equation. This is in accordance with the law of conservation of mass, which states that matter cannot be created or destroyed in an ordinary chemical reaction.

For instance, the equation in our exercise,
\[2 \mathrm{NaN}_3(\mathrm{s}) \rightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_2(\mathrm{g})\]
shows that two moles of solid sodium azide decompose to form two moles of solid sodium and three moles of nitrogen gas. Each side of the equation has an equal number of sodium and nitrogen atoms. The balancing of this equation is vital as it forms the foundation for stoichiometric conversions and calculations.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a fundamental concept in chemistry that allows us to convert between the mass of a substance and the amount in moles. Accurately calculating molar mass is necessary for stoichiometry, as it bridges the gap between the microscopic world (atoms, molecules) and the macroscopic (grams, liters) that we measure in the lab.

In our exercise, we determined the molar mass of NaN3 by summing the molar masses of its constituent atoms:
\[\text{Molar mass of } \mathrm{NaN}_3 = 1 \times 22.99 \text{ g/mol (sodium) } + 3 \times 14.01 \text{ g/mol (nitrogen) } = 65.01 \text{ g/mol}\].
Similarly, we found the molar mass of N2:
\[\text{Molar mass of } \mathrm{N}_2 = 2 \times 14.01 \text{ g/mol (nitrogen) } = 28.02 \text{ g/mol}\].
Knowing these values allows for the conversion from mass to moles, which is the next step in finding the amount of N2 produced.

Stoichiometric Conversions

Stoichiometric conversions are calculations that use the relationship between substances in a balanced chemical equation to determine the amount of products or reactants. These are fundamental to predictive chemistry—knowing what amount of a substance will be necessary or produced under given conditions.

In practice, once we've balanced the equation and found the molar masses, we perform a stoichiometric conversion to find out, for example, how many grams of nitrogen gas will be produced from a known mass of sodium azide. We do this by first converting the mass of NaN3 to moles, using its molar mass:
\[\text{Moles of } \mathrm{NaN}_3 = \frac{\text{Given mass of } \mathrm{NaN}_3}{65.01 \text{ g/mol}}\].
Then, applying the stoichiometric ratio from the balanced equation tells us how many moles of N2 are produced per mole of NaN3. Multiplying by the molar mass of N2 gives us the final mass of the gas produced:
\[\text{Mass of } \mathrm{N}_2 = \text{Moles of } \mathrm{N}_2 \times 28.02 \text{ g/mol}\].
The simplification of the relevant expressions is the final step to arrive at the practical, usable number that tells you the quantity of N2 you can expect from your starting quantity of NaN3. This is stoichiometry in action, a cornerstone for all quantitative work in chemistry.