Question

When nitroglycerin explodes, it decomposes to form carbon dioxide gas, nitrogen gas, oxygen gas, and water vapor. The balanoed equation is $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}(l) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) If \(1.00\) mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\) decomposes, how many moles of each gaseous product should form? (b) If \(2.50 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\) decomposes, how many moles of each gaseous product should form?

Short answer

(a) For 1.00 mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\), 3.00 mol \(\mathrm{CO}_{2}\), 1.50 mol \(\mathrm{N}_{2}\), 0.25 mol \(\mathrm{O}_{2}\), and 2.50 mol \(\mathrm{H}_{2} \mathrm{O}\) are formed. (b) For 2.50 mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\), 7.50 mol \(\mathrm{CO}_{2}\), 3.75 mol \(\mathrm{N}_{2}\), 0.625 mol \(\mathrm{O}_{2}\), and 6.25 mol \(\mathrm{H}_{2} \mathrm{O}\) are formed.

Step-by-step solution

Analyze the Balanced Chemical Equation

From the balanced chemical equation, \(4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{N}_{3}(l) \rightarrow 12 \mathrm{CO}_{2}(g)+ 6 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\), it can be inferred that: for every 4 moles of \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9}\mathrm{N}_{3}\) that decomposes, 12 moles of \(\mathrm{CO}_{2}\), 6 moles of \(\mathrm{N}_{2}\), 1 mole of \(\mathrm{O}_{2}\), and 10 moles of \(\mathrm{H}_{2} \mathrm{O}\) are formed.

Calculate Moles for Each Gas for 1.00 Mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\)

According to mole ratio from the balanced chemical equation, if \(1.00\) mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\) decomposes, then \(3.00\) mol \(\mathrm{CO}_{2}\), \(1.50\) mol \(\mathrm{N}_{2}\), \(0.25\) mol \(\mathrm{O}_{2}\), and \(2.50\) mol \(\mathrm{H}_{2} \mathrm{O}\) will be formed.

Calculate Moles for Each Gas for 2.50 Mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\)

According to mole ratio from the balanced chemical equation, if \(2.50\) mol \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{9} \mathrm{~N}_{3}\) decomposes, then \(7.50\) mol \(\mathrm{CO}_{2}\), \(3.75\) mol \(\mathrm{N}_{2}\), \(0.625\) mol \(\mathrm{O}_{2}\), and \(6.25\) mol \(\mathrm{H}_{2} \mathrm{O}\) will be formed.

Key concepts

Stoichiometry

Stoichiometry is a core concept in chemistry that involves the calculation of reactants and products in chemical reactions based on the balanced equations. It allows chemists to predict the quantities of substances consumed and produced in a given reaction. In our example with nitroglycerin, stoichiometry helps us establish a direct ratio between the amount of nitroglycerin that decomposes and the amounts of gases produced as a result.
Understanding stoichiometry is crucial because it provides the quantitative backbone for chemical reactions:

• It helps determine how much of each reactant is needed.
• Predicts the amount of products formed.
• Ensures optimal use of reactants, reducing waste and energy consumption.

To apply stoichiometry, one must refer to the coefficients in the balanced equation. For instance, the decomposition of nitroglycerin into gases requires analyzing how the four moles of nitroglycerin relate to the moles of CO\(_2\), N\(_2\), O\(_2\), and H\(_2\)O produced. This is achieved through mole ratios derived from the balanced chemical equation.

Balanced Equations

A balanced chemical equation is key to applying stoichiometry effectively. It ensures that the same number of each type of atom is present on both sides of the reaction. This reflects the principle of conservation of mass, where matter is neither created nor destroyed in a chemical reaction.
For instance, with nitroglycerin's decomposition reaction:

\[ 4 \mathrm{C}_3 \mathrm{H}_5 \mathrm{O}_9 \mathrm{N}_3 \rightarrow 12 \mathrm{CO}_2 + 6 \mathrm{N}_2 + \mathrm{O}_2 + 10 \mathrm{H}_2 \mathrm{O} \]

each coefficient (like "4" in front of nitroglycerin) is there to ensure that atoms like carbon, hydrogen, oxygen, and nitrogen are balanced. Without this balance, stoichiometric calculations would not provide meaningful predictions about reactant or product quantities.

• The equation must account for all atoms: if you start with 4 nitrogen atoms in the reactants, you must have 4 in the products.
• The balancing act often involves trial and error, making sure coefficients are the smallest whole numbers.
• Understanding this concept is foundational for predicting how much reactant or product one will have before or after a reaction.

For example, in balancing our given equation, these coefficients tell us precisely the proportion at which nitroglycerin decomposes into carbon dioxide, nitrogen, oxygen, and water vapor.

Mole Concept

The mole is a fundamental concept in chemistry that provides a bridge between the atomic scale and the macroscopic world we can measure. One mole of any substance contains Avogadro's number of entities, approximately 6.022 \(\times\) 10\(^{23}\). This conversion between atoms/molecules and grams is vital for quantifying reactions.
In the context of nitroglycerin decomposition:

• The balanced equation tells us that 4 moles of nitroglycerin produce specific moles of each gas.
• Using the mole concept, 1 mole of nitroglycerin links directly to the production of a certain number of moles of CO\(_2\), N\(_2\), O\(_2\), and H\(_2\)O
• We can calculate the exact number of moles of gas produced when a certain number of moles of nitroglycerin react. For example, decomposing 1 mole of nitroglycerin produces 3 moles of CO\(_2\), 1.5 moles of N\(_2\), 0.25 moles of O\(_2\), and 2.5 moles of H\(_2\)O.

It's crucial for students to understand that the mole provides a consistent method to convert between the mass of a substance and the number of particles involved, making stoichiometric calculations possible and practical.