Problem 10 - Chemistry The Molecular Science

Question

For this reaction, fill in the table with the indicated quantities for the balanced equation. $$4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ $$\begin{array}{|l|l|l|l|l|}\hline & \mathrm{NH}_{3} & \mathrm{O}_{2} & \mathrm{NO} & \mathrm{H}_{2} \mathrm{O} \\\\\hline \text { No. of molecules } & & & & \\ \hline \text { No. of atoms } & & & & \\\\\hline \begin{array}{l}\text { Amount of } \\\\\text { molecules }\end{array} & & & & \\\\\hline \text { Mass } & & & & \\\\\hline \begin{array}{l}\text { Total mass } \\\\\text { of reactants }\end{array} & & & \\\\\hline \begin{array}{l}\text { Total mass } \\\\\text { of products }\end{array} & & & & \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Each reactant/product has indicated numbers of molecules: NH₃ (4, 68 g), O₂ (5, 160 g), NO (4, 120 g), H₂O (6, 108 g). Total mass is conserved at 228 g for reactants and products.

Step 1: Identifying the Molecules

Each coefficient in the balanced chemical equation represents the number of molecules. From the equation, you know there are 4 NH₃, 5 O₂, 4 NO, and 6 H₂O molecules.

Step 2: Counting Total Atoms

Calculate the total number of atoms for each species. For NH₃: 4 molecules x 4 atoms/molecule (3 H + 1 N) = 16 atoms, for O₂: 5 molecules x 2 atoms/molecule = 10 atoms, for NO: 4 molecules x 2 atoms/molecule = 8 atoms, for H₂O: 6 molecules x 3 atoms/molecule (2 H + 1 O) = 18 atoms.

Step 3: Amount of Molecules

Here, the amount of molecules is the same as the number of molecules identified in Step 1. Fill in the table with 4 for NH₃, 5 for O₂, 4 for NO, and 6 for H₂O.

Step 4: Calculate the Mass

Use molar masses for calculation. Molar mass of NH₃ = 17 g/mol, O₂ = 32 g/mol, NO = 30 g/mol, and H₂O = 18 g/mol. Multiply the molar mass by the number of molecules (considering 1 mol = Avogadro's number of molecules): Mass of NH₃ = 4 x 17 g/mol = 68 g, Mass of O₂ = 5 x 32 g/mol = 160 g, Mass of NO = 4 x 30 g/mol = 120 g, Mass of H₂O = 6 x 18 g/mol = 108 g.

Step 5: Calculate Total Mass of Reactants

Add the masses of reactants: Total mass of NH₃ and O₂ = 68 g + 160 g = 228 g.

Step 6: Calculate Total Mass of Products

Add the masses of products: Total mass of NO and H₂O = 120 g + 108 g = 228 g, verifying the conservation of mass.

Key Concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It's about figuring out how much of each substance is involved. When you look at a balanced chemical equation, like the one given, the coefficients tell you the ratios of the molecules involved. In this example, for every 4 molecules of ammonia (\(\text{NH}_3\)), you need 5 molecules of oxygen (\(\text{O}_2\)) to produce 4 molecules of nitrogen monoxide (\(\text{NO}\)) and 6 molecules of water (\(\text{H}_2\text{O}\)). This relationship is crucial because it helps us predict how much reactant is needed to create a certain amount of product.

• 4 molecules of \(\text{NH}_3\) react with 5 molecules of \(\text{O}_2\)
• Produce 4 molecules of \(\text{NO}\) and 6 molecules of \(\text{H}_2\text{O}\)

This idea of stoichiometry helps chemists calculate how much of each reactant they need or how much product they can get with the given reactants. Knowing these relationships also helps ensure that there is no excess reactant left unused, making the reaction efficient.

Molar Mass Calculation

Molar mass is like a bridge between the atomic scale and the macroscale of grams. It allows us to translate the number of atoms or molecules into a mass that we can measure. To calculate the molar mass of a compound, you sum up the masses of each individual atom within a molecule according to their stoichiometric numbers.Consider ammonia (\(\text{NH}_3\)): - Nitrogen (N) has an atomic mass of approximately 14 g/mol. - Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are 3 hydrogens.Adding these gives \(17 \text{ g/mol}\) for ammonia.

• Multiply the molar mass by the number of molecules to find the total mass involved. For example, for ammonia: \[\text{Mass of NH}_3 = 4 \text{ molecules} \times 17 \text{ g/mol} = 68 \text{ g}\]
• Repeat this process for other substances, like oxygen, nitrogen monoxide, and water.

This calculation is necessary to accurately weigh out chemicals in the lab and ensures each side of the equation is balanced not just in numbers but in mass as well.

Conservation of Mass

The conservation of mass is a fundamental principle in chemistry that states that mass is neither created nor destroyed in a chemical reaction. This means the total mass of reactants is always equal to the total mass of products. In the given reaction, this principle is clearly demonstrated by the balanced chemical equation.When we calculated the masses:- Reactants: Total mass from ammonia (\(\text{NH}_3\)) and oxygen (\(\text{O}_2\)) was\[68 \text{ g} + 160 \text{ g} = 228 \text{ g}\]- Products: Total mass from nitrogen monoxide (\(\text{NO}\)) and water (\(\text{H}_2\text{O}\)) also totaled\[120 \text{ g} + 108 \text{ g} = 228 \text{ g}\]This simply confirms that the mass is conserved. Understanding this principle allows chemists to ensure that their equations are correctly balanced and to trust that the calculated masses and molar ratios will hold true in practical experiments. This is key when scaling reactions from a lab bench to industrial production and is fundamental in all branches of chemistry.