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

Hence, \(98.97 \textrm{ g}\) of \(\textrm{F}_2\) are needed to produce \(120 \textrm{ g}\) of \(\textrm{PF}_3\) with a \(78.1\%\) yield.

Step-by-step solution

Balance the chemical equation.

The given chemical equation is: \( \textrm{P}_4(s) + \textrm{F}_2(g) \longrightarrow \textrm{PF}_3(g) \) Balancing the chemical equation, we get: \( \textrm{P}_4(s) + 6\textrm{F}_2(g) \longrightarrow 4\textrm{PF}_3(g) \)

Obtain the stoichiometric ratio.

From the balanced chemical equation, the stoichiometric ratios are: 1 mol P4 : 6 mol F2 : 4 mol PF3

Calculate the moles of PF3 formed.

Given the mass of PF3 is 120 g, let's find the number of moles using the molar mass of PF3 (1P + 3F = 1 × 30.97 + 3 × 19.00 = 88.97 g/mol): Moles of PF3 = (mass of PF3) / (molar mass of PF3) moles of PF3 = (120 g) / (88.97 g/mol) = 1.349 mol

Use the stoichiometry to find the moles of F2 required.

From the stoichiometric ratio, 1 mol of P4 requires 6 mol of F2 to produce 4 mol of PF3. So, let's calculate the moles of F2 required to produce 1.349 mol of PF3: Moles of F2 needed = (1.349 mol PF3) × (6 mol F2 / 4 mol PF3) = 2.023 mol F2

Convert the moles of F2 to mass, considering the yield.

To find mass of F2 needed, we use the molar mass of F2 (2 × 19.00 g/mol = 38.00 g/mol) and the given yield (78.1%): Mass of F2 required = (moles of F2 needed) × (molar mass of F2) × (100 / yield) Mass of F2 required = (2.023 mol) × (38.00 g/mol) × (100 / 78.1) Mass of F2 required = 98.97 g Hence, 98.97 g of F2 are needed to produce 120 g of PF3 with a 78.1% yield.

Key concepts

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in stoichiometry. It ensures that the Law of Conservation of Mass is upheld, meaning that matter is neither created nor destroyed in a chemical reaction. To balance an equation, adjust the coefficients (the numbers in front of molecules) so that the number of atoms of each element is the same on both sides of the equation. In our example, the original unbalanced equation is:

• \( \textrm{P}_4(s) + \textrm{F}_2(g) \to \textrm{PF}_3(g) \)

The final balanced equation we obtain is:

• \( \textrm{P}_4(s) + 6\textrm{F}_2(g) \to 4\textrm{PF}_3(g) \)

Here, adding a 6 in front of \( \textrm{F}_2 \) and a 4 in front of \( \textrm{PF}_3 \) ensures that there are equal numbers of each atom on both sides.

Molar Mass Calculation

The concept of molar mass is crucial for converting between the mass of a substance and the number of moles. To find the molar mass, add together the atomic masses of all atoms in a compound. Say we want to calculate for \( \textrm{PF}_3 \);

• Atomic mass of P = 30.97 g/mol
• Atomic mass of F = 19.00 g/mol

Then,

• Molar mass of \( \textrm{PF}_3 = 1 \times 30.97 + 3 \times 19.00 = 88.97 \text{ g/mol} \)

This molar mass is used to convert grams of \( \textrm{PF}_3 \) into moles, a necessary step for analyzing the reaction quantitatively.

Limiting Reactants

A limiting reactant is the substance that is fully consumed first during a chemical reaction. It limits the amount of product formed. In order to determine which reactant is limiting, compare the number of moles of each substance to the stoichiometric coefficients in the balanced equation. Using the stoichiometric ratio from our balanced equation:

• 1 mol \( \textrm{P}_4 : 6 \text{ mol } \textrm{F}_2 : 4 \text{ mol } \textrm{PF}_3 \)

This suggests that for every 4 moles of \( \textrm{PF}_3 \), we need 6 moles of \( \textrm{F}_2 \). If the available moles of \( \textrm{F}_2 \) aren’t enough to fully react with \( \textrm{P}_4 \), it becomes the limiting reactant and determines the maximum yield.

Percentage Yield Calculations

The percentage yield shows how efficient a reaction is in producing the desired product compared to the theoretical maximum. It is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. In this case, the reaction has a 78.1% yield, meaning:

• Actual yield (from the problem) = 120 g of \( \textrm{PF}_3 \)

The strategic steps are:

• Calculate the theoretical yield having all reactants completely react (not shown here for brevity).
• Use the equation:\[\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]
• Solve for required initial masses accordingly, as done in our problem where 98.97 g of \( \textrm{F}_2 \) was calculated to achieve the 78.1% yield.

The percentage yield helps in understanding the practicality of the reaction outcomes in a real-world setting.