Problem 83 - 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 \(\mathrm{NaN}_{3}\) is produced.

Step 1: Calculate Molar Masses

First, calculate the molar masses of the reactants and the product we are interested in. - Molar mass of \(\mathrm{NaNO}_{3}\): \( 22.99 + 14.01 + 3(16.00) = 85.00 \, \mathrm{g/mol} \) - Molar mass of \(\mathrm{NaNH}_{2}\): \( 22.99 + 14.01 + 2(1.01) = 39.01 \, \mathrm{g/mol} \)- Molar mass of \(\mathrm{NaN}_{3}\): \( 22.99 + 3(14.01) = 65.02 \, \mathrm{g/mol} \)

Step 2: Convert Given Mass to Moles

Calculate the amount in moles for each reactant using their masses and molar masses:- Moles of \(\mathrm{NaNO}_{3}\): \( \frac{15.0 \, \mathrm{g}}{85.00 \, \mathrm{g/mol}} = 0.176 \, \mathrm{mol} \) - Moles of \(\mathrm{NaNH}_{2}\): \( \frac{15.0 \, \mathrm{g}}{39.01 \, \mathrm{g/mol}} = 0.384 \, \mathrm{mol} \)

Step 3: Determine Limiting Reagent

The balanced equation indicates a 1:3 molar ratio of \(\mathrm{NaNO}_{3}\) to \(\mathrm{NaNH}_{2}\). For 0.176 mol of \(\mathrm{NaNO}_{3}\), we need \(0.176 \times 3 = 0.528\) mol of \(\mathrm{NaNH}_{2}\). Since we only have 0.384 mol of \(\mathrm{NaNH}_{2}\), which is less than 0.528 mol, \(\mathrm{NaNH}_{2}\) is the limiting reagent.

Step 4: Calculate Theoretical Yield

Use the amount of the limiting reagent to find the amount of the product \(\mathrm{NaN}_{3}\):- According to the reaction, 3 mol of \(\mathrm{NaNH}_{2}\) gives 1 mol of \(\mathrm{NaN}_{3}\), hence, \(0.384 \div 3 = 0.128\) mol of \(\mathrm{NaN}_{3}\) can be produced.

Step 5: Convert Moles of Product to Mass

Convert the moles of \(\mathrm{NaN}_{3}\) back to grams using its molar mass:- Mass of \(\mathrm{NaN}_{3}\): \(0.128 \, \mathrm{mol} \times 65.02 \, \mathrm{g/mol} = 8.32256 \, \mathrm{g} \). Thus, approximately 8.32 g of \(\mathrm{NaN}_{3}\) is produced.

Key Concepts

Limiting Reagent

In chemical reactions, the limiting reagent is the substance that gets used up first, limiting the amount of product that can be formed. Understanding which reactant is the limiting reagent is crucial. It determines when the reaction will stop. It's like baking cookies; if you run out of dough, you can't make more cookies no matter how much frosting you have left.
In the given reaction, we start by comparing the mole ratio of available reactants to the mole ratio required by the balanced equation. The equation \[\mathrm{NaNO}_3 + 3 \mathrm{NaNH}_2 \rightarrow \mathrm{NaN}_3 + 3 \mathrm{NaOH} + \mathrm{NH}_3\]shows that 1 mole of \(\mathrm{NaNO}_3\) needs 3 moles of \(\mathrm{NaNH}_2\). In this scenario, \(\mathrm{NaNH}_2\) ran out first because even though we had 0.384 mol of \(\mathrm{NaNH}_2\), we would need 0.528 mol to completely react with 0.176 mol of \(\mathrm{NaNO}_3\). Therefore, \(\mathrm{NaNH}_2\) is the limiting reagent.

Theoretical Yield

Theoretical yield is the maximum amount of product that can be produced in a chemical reaction, based on the amount of limiting reagent. It assumes perfect conditions where no material is lost or wasted.
The calculation of theoretical yield often involves first identifying the limiting reagent. As we saw previously, \(\mathrm{NaNH}_2\) is the limiting reagent. Once identified, we use stoichiometry to determine how much product can be produced. According to the balanced chemical equation, 3 moles of \(\mathrm{NaNH}_2\) produce 1 mole of \(\mathrm{NaN}_3\). Hence, from 0.384 mol of \(\mathrm{NaNH}_2\), we can theoretically obtain 0.128 mol of \(\mathrm{NaN}_3\).
This reflects ideal conditions. In practice, actual yield is often less due to various inefficiencies or side reactions.

Molar Mass Calculation

Molar mass is a central concept in stoichiometry, acting as the bridge between mass and moles. It helps us convert a given amount of a substance from grams to moles and vice versa.
To calculate the molar mass of a compound, sum up the atomic masses of all atoms in a molecule. For instance:

• The molar mass of \(\mathrm{NaNO}_3\) is obtained by adding the masses of one sodium (Na) atom, one nitrogen (N) atom, and three oxygen (O) atoms:
• \[22.99 + 14.01 + 3(16.00) = 85.00 \, \mathrm{g/mol}\]
• Similarly, calculate \(\mathrm{NaNH}_2\): \[22.99 + 14.01 + 2(1.01) = 39.01 \, \mathrm{g/mol}\]
• Finally, for \(\mathrm{NaN}_3\) use: \[22.99 + 3(14.01) = 65.02 \, \mathrm{g/mol}\]

Molar masses aid in translating between the mass of a sample and the number of moles it contains, making stoichiometric calculations possible.

Chemical Reactions

Chemical reactions occur when substances interact and transform into different substances. These are represented by chemical equations that show the reactants turning into products.
Each reaction follows specific stoichiometric patterns, wherein balanced chemical equations reflect the conservation of mass and charge. For the reaction \[\mathrm{NaNO}_3 + 3 \mathrm{NaNH}_2 \rightarrow \mathrm{NaN}_3 + 3 \mathrm{NaOH} + \mathrm{NH}_3\], it is crucial to ensure that the number of each type of atom is the same on both sides of the equation.
Understanding stoichiometry lets us predict amounts of reactants and products involved in the chemical reaction. It's not just about knowing what we start and end with, but also about understanding the interrelationships between substances and the efficient, mathematical path from start to finish. This fundamental concept ensures precise calculations and understanding of chemical transformations.