Problem 56 - Chemistry

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 \(\mathrm{P}_{4} \mathrm{O}_{10}\) are reacted with an excess of \(\mathrm{HNO}_{3}\). (Hint: One of the products is \(\mathrm{HPO}_{3}\).)

Step-by-Step Solution

Answer

The balanced equation for the reaction is \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\) and the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(60.3 \mathrm{g}\)

Step 1: Write Unbalanced equation

Start by writing the unbalanced chemical equation. This would be \(\mathrm{P}_{4} \mathrm{O}_{10} + \mathrm{HNO}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5} + \mathrm{HPO}_{3}\)

Step 2: Balance the equation

Balance the chemical equation. When balanced, it gives: \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\)

Step 3: Calculate molar mass of P4O10 and N2O5

Calculate the molar mass of \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{N}_{2} \mathrm{O}_{5}\). The molar mass of \(\mathrm{P}_{4} \mathrm{O}_{10} = 283.9 \mathrm{~g/mol}\) and of \(\mathrm{N}_{2} \mathrm{O}_{5} = 108.01 \mathrm{~g/mol}\)

Step 4: Use stoichiometry to calculate theoretical yield

According to the balanced equation, 1 mol of \(\mathrm{P}_{4} \mathrm{O}_{10}\) forms 2 mol \(\mathrm{N}_{2} \mathrm{O}_{5}\). Convert given mass of \(\mathrm{P}_{4} \mathrm{O}_{10}\) to moles by dividing by its molar mass, then multiply by 2 to find moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\), and finally convert back to grams by multiplying with its molar mass which gives the theoretical yield. \((79.4 \mathrm{g} \div 283.9 \mathrm{g/mol}) \times 2 \times 108.01 \mathrm{g/mol} = 60.3 \mathrm{g}\)

Step 5: Conclusion

The balanced chemical equation for the above reaction is \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\) and the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\) when \(79.4 \mathrm{g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) are reacted with an excess of \(\mathrm{HNO}_{3}\) is \(60.3 \mathrm{g}\)

Key Concepts

Balanced Chemical Equation

When approaching chemical reactions, balancing the equation is an essential first step. This ensures that the number of atoms for each element is conserved throughout the reaction.
In the given reaction between \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{HNO}_{3}\), we start by writing the unbalanced equation: \\(\mathrm{P}_{4} \mathrm{O}_{10} + \mathrm{HNO}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5} + \mathrm{HPO}_{3}\).
Next, we apply the law of conservation of mass to adjust coefficients until each type of atom has the same number on both sides of the equation.The balanced equation becomes: \\(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\).

• The coefficient 12 in front of \(\mathrm{HNO}_{3}\) indicates that 12 moles of \(\mathrm{HNO}_{3}\) are required for a complete reaction.
• A 2 in front of \(\mathrm{N}_{2} \mathrm{O}_{5}\) shows the formation of two moles of this compound.
• This balance is crucial for accurate stoichiometric calculations in the next steps.

Theoretical Yield

Understanding theoretical yield is vital in chemistry as it predicts the maximum amount of product formed during a reaction, assuming complete conversion of reactants.
In our example, with the balanced equation: \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\), we're asked to find the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\).
This necessitates using the given mass of \(\mathrm{P}_{4} \mathrm{O}_{10}\) (79.4 g). The theoretical yield calculation involves:

• Converting the mass of \(\mathrm{P}_{4} \mathrm{O}_{10}\) to moles, using its molar mass.
• Determining moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) produced based on the stoichiometry from the balanced equation.
• Converting moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) back to grams using its molar mass to find the theoretical yield.

In this case, the calculation shows a theoretical yield of 60.3 g of \(\mathrm{N}_{2} \mathrm{O}_{5}\), reflecting the maximum possible quantity achievable under ideal conditions.

Molar Mass Calculation

Calculating molar mass is a fundamental skill in chemical stoichiometry, required for converting between mass and moles.
Molar mass is the sum of atomic masses of all atoms in a molecule, expressed in g/mol.
In our reaction, we calculate it for \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{N}_{2} \mathrm{O}_{5}\) to facilitate stoichiometric calculations:

• For \(\mathrm{P}_{4} \mathrm{O}_{10}\): Add the atomic masses of 4 phosphorus atoms and 10 oxygen atoms: \(4 \times 30.97 + 10 \times 16\), resulting in a molar mass of 283.9 g/mol.
• For \(\mathrm{N}_{2} \mathrm{O}_{5}\): Combine the masses of 2 nitrogen and 5 oxygen atoms: \(2 \times 14.01 + 5 \times 16\), equaling 108.01 g/mol.

These values enable you to effectively convert grams to moles, and vice versa, which is pivotal for determining the quantities in chemical reactions and calculating the theoretical yield accurately.