Problem 85 - Chemistry and Chemical Reactivity

Question

Sodium azide, an explosive chemical used in automobile airbags, is made by the following reaction: $$\mathrm{NaNO}_{3}+3 \mathrm{NaNH}_{2} \rightarrow \mathrm{NaN}_{3}+3 \mathrm{NaOH}+\mathrm{NH}_{3}$$ If you combine \(15.0 \mathrm{g}\) of \(\mathrm{NaNO}_{3}\) with \(15.0 \mathrm{g}\) of \(\mathrm{NaNH}_{2},\) what mass of \(\mathrm{NaN}_{3}\) is produced?

Step-by-Step Solution

Answer

8.32 g of NaN₃ is produced.

Step 1: Calculate Molar Masses

First, we need to find the molar masses of the substances involved. The molar mass of \(\mathrm{NaNO}_3\) is \(85 \text{ g/mol}\) (\(\text{Na} = 23 \text{ g/mol}, \text{N} = 14 \text{ g/mol}, \text{O}_3 = 3 \times 16 \text{ g/mol}\)). The molar mass of \(\mathrm{NaNH}_2\) is \(39 \text{ g/mol}\) (\(\text{Na} = 23 \text{ g/mol}, \text{N} = 14 \text{ g/mol}, \text{H}_2 = 2 \times 1 \text{ g/mol}\)). The molar mass of \(\mathrm{NaN}_3\) is \(65 \text{ g/mol}\) (\(\text{Na} = 23 \text{ g/mol}, \text{N}_3 = 3 \times 14 \text{ g/mol}\)).

Step 2: Convert Mass to Moles

Convert the given masses of \(\mathrm{NaNO}_3\) and \(\mathrm{NaNH}_2\) to moles. \(\text{Moles of NaNO}_3 = \frac{15.0 \text{ g}}{85 \text{ g/mol}} = 0.176 \text{ mol}\). \(\text{Moles of NaNH}_2 = \frac{15.0 \text{ g}}{39 \text{ g/mol}} = 0.385 \text{ mol}\).

Step 3: Determine Limiting Reagent

According to the balanced equation, 1 mole of \(\mathrm{NaNO}_3\) reacts with 3 moles of \(\mathrm{NaNH}_2\). Therefore, for 0.176 moles of \(\mathrm{NaNO}_3\), we would need 0.528 moles of \(\mathrm{NaNH}_2\). Since only 0.385 moles of \(\mathrm{NaNH}_2\) are available, \(\mathrm{NaNH}_2\) is the limiting reagent.

Step 4: Calculate Moles of Product

Using the limiting reagent, convert moles of \(\mathrm{NaNH}_2\) to moles of \(\mathrm{NaN}_3\). For 3 moles of \(\mathrm{NaNH}_2\), 1 mole of \(\mathrm{NaN}_3\) is formed. Therefore, \(\text{Moles of NaN}_3 = \frac{0.385 \text{ mol NaNH}_2}{3} = 0.128 \text{ mol NaN}_3\).

Step 5: Convert Moles to Mass

Convert the moles of \(\mathrm{NaN}_3\) to grams. \(\text{Mass of NaN}_3 = 0.128 \text{ mol} \times 65 \text{ g/mol} = 8.32 \text{ g}\).

Key Concepts

Limiting Reagents

In stoichiometry, the concept of limiting reagents helps us understand which reactant will run out first during a chemical reaction. This reactant determines the maximum amount of product that can be formed. A limiting reagent is the substance that is fully consumed first, stopping the reaction from continuing because there is nothing left to react with the other reagent.

To find the limiting reagent, follow these steps:

• Convert the given masses of reactants to moles using their molar masses.
• Use the coefficients from the balanced chemical equation to determine the exact proportions needed for the reaction.
• Calculate how many moles of each reactant are required and identify which reactant has less than what is needed according to the stoichiometric ratio.

In this exercise, \(\text{NaNH}_2\) was found to be the limiting reagent because it does not meet the stoichiometric requirement when compared to \(\text{NaNO}_3\) for the reaction to complete.
Understanding limiting reagents is crucial, as it ensures that you can accurately predict the quantities of products formed in reactions involving multiple reactants.

Molar Mass Calculations

Molar mass is a key concept in stoichiometry as it allows us to convert between grams and moles. This conversion is necessary because chemical reactions occur at the molecular level, where substances react in moles rather than grams. To solve stoichiometry problems, calculating molar mass properly is crucial.

Here's how to calculate molar mass for a compound:

• Identify the elements present in the compound and their atomic masses from the periodic table.
• Multiply the atomic mass of each element by the number of times the element appears in the compound.
• Add up all these values to get the total molar mass of the compound.

For example, the molar mass of \(\text{NaNO}_3\) is calculated as follows: \(\text{Na} = 23 \, \text{g/mol}, \, \text{N} = 14 \, \text{g/mol}, \, \text{O}_3 = 3 \times 16 \, \text{g/mol}\). Hence, the molar mass of \(\text{NaNO}_3\) is \(85 \, \text{g/mol}\).
By converting mass to moles, we can use stoichiometry to predict reactants' outcomes and understand the dynamics of chemical reactions.

Chemical Equations

Chemical equations provide a detailed representation of chemical reactions indicating the reactants, products, and their proportions. Understanding chemical equations is fundamental to solving stoichiometry problems efficiently. Here are some key points:

• A balanced chemical equation means that the number of atoms for each element is the same on both sides of the equation.
• The coefficients in the equations show the mole ratio of reactants and products, which is essential in deciding the amount of product when starting with given reactants.
• Chemical equations are crucial in determining theoretical yields and identifying limiting reagents when reactants are not in perfect proportion.

Consider the equation \(\text{NaNO}_3 + 3\text{NaNH}_2 \rightarrow \text{NaN}_3 + 3\text{NaOH} + \text{NH}_3\). This balance shows that for every 1 mole of \(\text{NaNO}_3\), 3 moles of \(\text{NaNH}_2\) are required to produce 1 mole of \(\text{NaN}_3\).Understanding chemical equations allows students to connect the theoretical aspects of chemistry with practical, real-world applications in laboratory or industrial settings.