Problem 55
Question
Sodium azide, the explosive chemical used in automobile airbags, is made by the following reaction: $$ \mathrm{NaNO}_{3}+3 \mathrm{NaNH}_{2} \longrightarrow \mathrm{NaN}_{3}+3 \mathrm{NaOH}+\mathrm{NH}_{3} $$ If you combine \(15.0 \mathrm{g}\) of \(\mathrm{NaNO}_{3}(85.0 \mathrm{g} / \mathrm{mol})\) with \(15.0 \mathrm{g}\) of \(\mathrm{NaNH}_{2},\) what mass of \(\mathrm{NaN}_{3}\) is produced?
Step-by-Step Solution
Verified Answer
Approximately 8.32 g of \( \mathrm{NaN}_{3} \) is produced.
1Step 1: Calculate moles of NaNO3
First, determine the moles of \( \mathrm{NaNO}_{3} \) by using its given mass and molar mass. \[ \text{Moles of } \mathrm{NaNO}_{3} = \frac{15.0 \text{ g}}{85.0 \text{ g/mol}} \approx 0.176 \text{ moles} \]
2Step 2: Calculate moles of NaNH2
Next, calculate the moles of \( \mathrm{NaNH}_{2} \) using its mass and molar mass. The molar mass of \( \mathrm{NaNH}_{2} \) is \(39.0 \text{ g/mol} \). Thus, \[ \text{Moles of } \mathrm{NaNH}_{2} = \frac{15.0 \text{ g}}{39.0 \text{ g/mol}} \approx 0.385 \text{ moles} \]
3Step 3: Identify the limiting reactant
According to the balanced equation, 1 mole of \( \mathrm{NaNO}_{3} \) reacts with 3 moles of \( \mathrm{NaNH}_{2} \). Check which is the limiting reactant: \[ \text{Required moles of } \mathrm{NaNH}_{2} = 3 \times 0.176 = 0.528 \text{ moles} \] Since we only have 0.385 moles of \( \mathrm{NaNH}_{2} \), it is the limiting reactant.
4Step 4: Calculate moles of NaN3 produced
From the stoichiometry of the reaction, 3 moles of \( \mathrm{NaNH}_{2} \) produce 1 mole of \( \mathrm{NaN}_{3} \). Use the limiting reactant to find the moles of \( \mathrm{NaN}_{3} \) produced: \[ \text{Moles of } \mathrm{NaN}_{3} = \frac{0.385}{3} = 0.128 \text{ moles} \]
5Step 5: Calculate mass of NaN3 produced
Finally, convert the moles of \( \mathrm{NaN}_{3} \) to grams using its molar mass (65.0 g/mol): \[ \text{Mass of } \mathrm{NaN}_{3} = 0.128 \times 65.0 = 8.32 \text{ g} \]
Key Concepts
StoichiometryMolar MassChemical Reaction
Stoichiometry
Stoichiometry is a key concept in chemistry that involves the calculation of reactants and products in chemical reactions. It relies on the balanced chemical equation, which provides a recipe-like guide to how molecules interact.
In the original exercise, this concept is used to determine how much product, specifically sodium azide (\(\text{NaN}_3\), can be produced from given amounts of reactants like sodium nitrate (\(\text{NaNO}_3\)) and sodium amide (\(\text{NaNH}_2\)).
To use stoichiometry, first ensure the chemical equation is balanced, showing equal numbers of each type of atom on both sides of the equation. For example, in the reaction:
\( \text{NaNO}_3 + 3\text{NaNH}_2 \rightarrow \text{NaN}_3 + 3\text{NaOH} + \text{NH}_3 \)
This equation tells us that one molecule of \(\text{NaNO}_3\) reacts with three molecules of \(\text{NaNH}_2\) to form one molecule of \(\text{NaN}_3\). This ratio is critical for calculating the "limiting reactant" and determining how much product is produced.
Stoichiometry is vital as it ensures chemicals are used efficiently and safely, minimizing waste and identifying the limiting factor in reactions.
In the original exercise, this concept is used to determine how much product, specifically sodium azide (\(\text{NaN}_3\), can be produced from given amounts of reactants like sodium nitrate (\(\text{NaNO}_3\)) and sodium amide (\(\text{NaNH}_2\)).
To use stoichiometry, first ensure the chemical equation is balanced, showing equal numbers of each type of atom on both sides of the equation. For example, in the reaction:
\( \text{NaNO}_3 + 3\text{NaNH}_2 \rightarrow \text{NaN}_3 + 3\text{NaOH} + \text{NH}_3 \)
This equation tells us that one molecule of \(\text{NaNO}_3\) reacts with three molecules of \(\text{NaNH}_2\) to form one molecule of \(\text{NaN}_3\). This ratio is critical for calculating the "limiting reactant" and determining how much product is produced.
Stoichiometry is vital as it ensures chemicals are used efficiently and safely, minimizing waste and identifying the limiting factor in reactions.
Molar Mass
Molar mass is an essential concept in chemistry that helps convert between the mass of a substance and the number of moles, which represent the amount of a chemical substance. It is expressed in grams per mole (g/mol).
In chemical calculations, such as determining the limiting reactant or product formation like in our exercise, molar mass allows us to relate the mass of reactants to the amounts needed according to the coefficients in the balanced equation.
In chemical calculations, such as determining the limiting reactant or product formation like in our exercise, molar mass allows us to relate the mass of reactants to the amounts needed according to the coefficients in the balanced equation.
- For \(\text{NaNO}_3\), the molar mass is 85.0 g/mol, letting us calculate moles from mass:
- For \(\text{NaNH}_2\), the molar mass is 39.0 g/mol.
- As for our product, \(\text{NaN}_3\), its molar mass is 65.0 g/mol.
Chemical Reaction
Chemical reactions involve the transformation of reactants into products. A reaction equation provides a symbolic representation of this process, as seen with sodium azide production:
\(\text{NaNO}_3 + 3\text{NaNH}_2 \rightarrow \text{NaN}_3 + 3\text{NaOH} + \text{NH}_3\)
During this transformation, bonds are broken in the reactants and new ones are formed in the products.
In these equations:
Understanding reactions is not only the key to predicting the amounts of products created but also for recognizing energetic and chemical changes accompanying these interactions. Thus, mastering chemical reactions, as shown in this exercise, is central to grasping broader chemical principles.
\(\text{NaNO}_3 + 3\text{NaNH}_2 \rightarrow \text{NaN}_3 + 3\text{NaOH} + \text{NH}_3\)
During this transformation, bonds are broken in the reactants and new ones are formed in the products.
In these equations:
- Everything on the left side of the arrow are reactants, the starting substances in the reaction.
- The substances on the right side, known as products, are the compounds formed from the reaction.
Understanding reactions is not only the key to predicting the amounts of products created but also for recognizing energetic and chemical changes accompanying these interactions. Thus, mastering chemical reactions, as shown in this exercise, is central to grasping broader chemical principles.
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