Question

Dinitrogen pentoxide is a product of the reaction between \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{HNO}_{3}\). Write a balanced equation for this reaction. Calculate the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\) if \(79.4 \mathrm{~g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) is combined with an excess of \(\mathrm{HNO}_{3}\). (Hint: One of the products is \(\mathrm{HPO}_{3}\).)

Short answer

The theoretical yield of \( \mathrm{N}_{2} \mathrm{O}_{5} \) is 60.49 g.

Step-by-step solution

Write the Unbalanced Chemical Equation

The reaction involves \( \mathrm{P}_{4}\mathrm{O}_{10} \) and \( \mathrm{HNO}_{3} \). Given the hint, one of the products is \( \mathrm{HPO}_{3} \), so the unbalanced equation is: \[ \mathrm{P}_{4}\mathrm{O}_{10} + \mathrm{HNO}_{3} \rightarrow \mathrm{N}_{2}\mathrm{O}_{5} + \mathrm{HPO}_{3} \]

Balance the Chemical Equation

To balance the equation, ensure that the number of each type of atom is the same on both sides:\[ 2\mathrm{P}_{4}\mathrm{O}_{10} + 8\mathrm{HNO}_{3} \rightarrow 4\mathrm{N}_{2}\mathrm{O}_{5} + 8\mathrm{HPO}_{3} \] This equation is balanced with 2 phosphorus, 4 nitrogen, and 10 oxygen atoms on each side.

Calculate the Molar Mass of \( \mathrm{P}_{4} \mathrm{O}_{10} \)

The molar mass of \( \mathrm{P}_{4}\mathrm{O}_{10} \) is calculated by adding the masses of each element: \[ 4 \times 30.97 \, \text{g/mol (P)} + 10 \times 16.00 \, \text{g/mol (O)} = 283.88 \, \text{g/mol} \]

Calculate Moles of \( \mathrm{P}_{4} \mathrm{O}_{10} \)

Using the mass provided (79.4 g) and the molar mass, calculate the moles of \( \mathrm{P}_{4}\mathrm{O}_{10} \):\[ \text{Moles of } \mathrm{P}_{4}\mathrm{O}_{10} = \frac{79.4 \, \text{g}}{283.88 \, \text{g/mol}} \approx 0.28 \, \text{mol} \]

Determine Theoretical Yield of \( \mathrm{N}_{2}\mathrm{O}_{5} \)

Using the balanced equation, \(2\mathrm{P}_{4}\mathrm{O}_{10} \) produces \(4\mathrm{N}_{2}\mathrm{O}_{5} \). Thus, 1 mole of \( \mathrm{P}_{4}\mathrm{O}_{10} \) produces 2 moles of \( \mathrm{N}_{2}\mathrm{O}_{5} \).For 0.28 moles of \( \mathrm{P}_{4}\mathrm{O}_{10} \), we get:\[ 0.28 \, \text{mol} \times 2 = 0.56 \, \text{mol of } \mathrm{N}_{2}\mathrm{O}_{5} \]

Calculate the Mass of \( \mathrm{N}_{2}\mathrm{O}_{5} \)

Calculate the molar mass of \( \mathrm{N}_{2}\mathrm{O}_{5} \):\[ 2 \times 14.01 \, \text{g/mol (N)} + 5 \times 16.00 \, \text{g/mol (O)} = 108.02 \, \text{g/mol} \]Then, calculate the mass from moles:\[ 0.56 \, \text{mol} \times 108.02 \, \text{g/mol} = 60.49 \, \text{g} \]

Final Result

The theoretical yield of \( \mathrm{N}_{2} \mathrm{O}_{5} \) from 79.4 g of \( \mathrm{P}_{4}\mathrm{O}_{10} \) with excess \( \mathrm{HNO}_{3} \) is approximately 60.49 g.

Key concepts

Stoichiometry: Understanding Reactant and Product Relationships

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It is crucial for determining how much of each substance is needed or produced. By using stoichiometry, we can predict the amount of products formed from a given amount of reactants.
This involves the use of balanced chemical equations, which represent the conservation of mass in a reaction. The coefficients in the equations indicate the molar ratios of the substances involved. It’s like a recipe! If you have one egg and need two for a recipe, you know you can't get a full serving. Stoichiometry works in the same way by ensuring you have the right proportions.
In our exercise, it's essential to understand the stoichiometric relationship between \( \mathrm{P}_{4}\mathrm{O}_{10} \) and \( \mathrm{N}_{2}\mathrm{O}_{5} \). Every mole of \( \mathrm{P}_{4}\mathrm{O}_{10} \) delivers two moles of \( \mathrm{N}_{2}\mathrm{O}_{5} \) according to the balanced equation. This proportionality lets you calculate the theoretical yield.

Theoretical Yield: Predicting Maximum Product Formation

Theoretical yield refers to the maximum amount of product that can be formed in a chemical reaction, assuming perfect conversion of reactants without any losses or side reactions. It is an ideal value often calculated to assess the efficiency of a reaction.
Using stoichiometry, you can calculate the theoretical yield by applying the molar ratios derived from the balanced equation. In the example exercise, \( 0.28 \) moles of \( \mathrm{P}_{4}\mathrm{O}_{10} \) can theoretically produce \( 0.56 \) moles of \( \mathrm{N}_{2}\mathrm{O}_{5} \). By converting this into grams, considering the molar mass of \( \mathrm{N}_{2}\mathrm{O}_{5} \), the yield is found to be \( 60.49 \) grams.
Bear in mind that the theoretical yield is rarely achieved in practice due to factors like incomplete reactions, practical losses during the experiment, or side reactions.

Balancing Equations: Ensuring Atom Conservation

Balancing chemical equations is fundamental in chemistry as it ensures the principle of conservation of mass is maintained. This principle states that matter can neither be created nor destroyed in a chemical reaction, meaning the mass of reactants must equal the mass of products.
To balance an equation, adjust the coefficients before each reactant and product so the number of atoms of each element is equal on both sides of the equation. It isn't about changing the substances, merely their proportions.
For our given reaction, we end up with balanced terms: \[ 2\mathrm{P}_{4}\mathrm{O}_{10} + 8\mathrm{HNO}_{3} \rightarrow 4\mathrm{N}_{2}\mathrm{O}_{5} + 8\mathrm{HPO}_{3} \]
This equation confirms that each element’s quantity is the same on both sides. It ensures that when we use stoichiometry, our mole ratios are accurate for calculating the theoretical yield.

Molar Mass Calculation: Converting Between Moles and Grams

Molar mass is the weight of one mole of any given element or compound and is significant in converting between the number of moles and the weight of a substance in grams.
Calculate the molar mass by summing up the atomic masses of all atoms present in the formula, measured in grams per mole (g/mol). For instance, the molar mass of \( \mathrm{P}_{4}\mathrm{O}_{10} \) is determined by adding \( 4 \times 30.97 \text{ (g/mol for P)} + 10 \times 16.00 \text{ (g/mol for O)} = 283.88 \text{ g/mol} \).
In the exercise, knowing the molar mass of \( \mathrm{P}_{4}\mathrm{O}_{10} \) allows us to convert 79.4 grams into moles, facilitating the stoichiometric calculations. This conversion is critical in progressing from mass-based measurements in a laboratory setting to mole-based stoichiometric calculations.