Question

Consider the following unbalanced reaction: $$\mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g)$$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

Short answer

In conclusion, \(99.52\:g\) of F₂ is needed to produce \(120\:g\) of PF₃ with a \(78.1\%\) yield.

Step-by-step solution

Balancing the chemical equation

Balance the given chemical equation by adding coefficients: \(P_4(s) + 6F_2(g) \longrightarrow 4PF_3(g)\) This balanced equation implies that one mole of P₄ reacts with 6 moles of F₂ to produce 4 moles of PF₃.

Determine the number of moles of PF₃

We are given that we want to produce 120 g of PF₃. First, we need to find the molar mass of PF₃: Molar mass of PF₃ = (mass of P) + 3 × (mass of F) = 30.97 g/mol + 3 × 19.00 g/mol = 87.97 g/mol. Now calculate the number of moles of PF₃: Number of moles = mass of product ÷ molar mass = 120 g ÷ 87.97 g/mol ≈ 1.364 moles of PF₃.

Determine the theoretical yield

The reaction has 78.1% yield, therefore: Actual yield = Theoretical yield × Percentage yield ÷ 100 Theoretical yield = Actual yield ÷ (Percentage yield ÷ 100) = 1.364 moles ÷ (78.1 ÷ 100) ≈ 1.746 moles of PF₃.

Find the moles of F₂ needed

Now, using the balanced equation (which states that 4 moles of PF₃ are produced by 6 moles of F₂), we can find the moles of F₂ needed by setting up a proportion: \(\frac{moles\:of\:F_2}{moles\:of\:PF_3} = \frac{6}{4}\) Moles of F₂ = (6/4) × Theoretical yield of PF₃ = (6/4) × 1.746 moles ≈ 2.619 moles of F₂

Convert moles of F₂ to mass of F₂

Finally, we determine the mass of F₂ needed. The molar mass of F₂ = 2 × mass of F = 2 × 19.00 g/mol = 38.00 g/mol. Mass of F₂ = moles of F₂ × molar mass of F₂ = 2.619 moles × 38.00 g/mol ≈ 99.52 g In conclusion, 99.52 g of F₂ is needed to produce 120 g of PF₃ with a 78.1% yield.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry. It ensures that the same number of atoms of each element are present on both sides of the reaction, adhering to the law of conservation of mass. In the given exercise, the unbalanced equation \( P_4(s) + F_2(g) \rightarrow PF_3(g) \) was balanced to \( P_4(s) + 6F_2(g) \rightarrow 4PF_3(g) \). This indicates that four phosphorus atoms from a single molecule of tetraphosphorus react with twelve fluorine atoms from six molecules of fluorine gas to produce four molecules of phosphorus trifluoride. Without this step, it would be impossible to accurately determine how much of each reactant is needed or predict the amount of product formed.

• To balance an equation, start by counting the number of atoms of each element on both sides of the reaction.
• Adjust the coefficients - the numbers in front of the compounds - to achieve the same number of atoms of each element on both sides.
• Bear in mind that the coefficients represent the smallest whole number ratio of reactants and products in the reaction.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced in a chemical reaction based on stoichiometry, assuming that every atom of the reactants converts to the product. It's a useful concept for predicting the maximum possible output under ideal conditions. In the context of the exercise, to find the theoretical yield, the actual yield of phosphorus trifluoride (120 g) was first taken for which the number of moles needed to be calculated.However, since reactions are rarely perfect in reality, the theoretical yield isn't the amount you can expect to get in an actual experiment. This is where the percentage yield comes into play, allowing you to adjust the theoretical yield to reflect the real-world efficiency of the reaction.

Understanding the relationship between actual, theoretical, and percentage yield is critical for planning and executing chemical syntheses and for economic considerations in industrial chemistry.

Percentage Yield

Percentage yield is a measure of an experiment's efficiency and is calculated by dividing the actual yield of a product by its theoretical yield, then multiplying by 100. It reflects how close the actual experiment was to the ideal conditions assumed in stoichiometric calculations. In our example, the reaction's percentage yield is given as 78.1%. This information is essential for determining the practical quantities of reactants needed.

Calculation of Theoretical Yield

Given the actual yield and the percentage yield, you can rearrange the percentage yield formula to calculate the theoretical yield: \(\text{Theoretical yield} = \frac{\text{Actual yield}}{\text{Percentage yield} \div 100}\).Using this adjusted value allows chemists to understand how much they should realistically expect to produce, thus refining the amounts of reactants required for a given amount of product. When planning a reaction, knowing the percentage yield is crucial for both economic and environmental reasons.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a determinant factor in converting between the mass of a substance and the number of moles. For stoichiometric calculations, you'll often need to calculate the molar masses of reactants and products, as seen in our exercise where the molar mass of phosphorus trifluoride (PF₃) had to be determined to convert grams to moles.To find the molar mass, sum the masses of all atoms within a molecule. The molar mass serves as a conversion factor in stoichiometry for relating the mass of a substance to the number of moles and vice versa. Accurate determination of molar masses is crucial for chemists to precisely measure the correct amounts of materials needed for reactions and to interpret results from experimental procedures.