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

99.25 g of F2 is needed to produce 120 g of PF3 with a 78.1% reaction yield.

Step-by-step solution

Balance the chemical equation

First, let's balance the given chemical equation: \(P_4(s) + 6F_2(g) \longrightarrow 4PF_3(g)\)

Calculate the amount of PF3 in moles

To calculate the amount of PF3 produced, we will convert the given mass (120 g) to moles. The molar mass of PF3 can be found using the periodic table: \(1P: 30.97g/mol\) \(3F: 19.00g/mol * 3 = 57.00g/mol\) Adding these together, the molar mass of PF3 is 87.97 g/mol. Now, let's convert the given mass of PF3 to moles: \(\frac{120\,\mathrm{g}}{1} * \frac{1\,\mathrm{mol\,PF_3}}{87.97\,\mathrm{g}} = 1.36\,\mathrm{mol\,PF_3}\)

Determine the amount of F2 required

Utilizing the stoichiometric coefficients, we can determine how many moles of F2 are required to produce the calculated moles of PF3. The mole ratio of F2 to PF3 is 6:4, which simplifies to 3:2 \(\frac{1.36\,\mathrm{mol\,PF_3}}{1} * \frac{3\,\mathrm{mol\,F_2}}{2\,\mathrm{mol\,PF_3}} = 2.04\,\mathrm{mol\,F_2}\)

Convert moles of F2 to mass

Now we need to convert the moles of F2 required into mass. The molar mass of F2 is: \(2F: 19.00g/mol * 2 = 38.00g/mol\) Converting moles of F2 to mass: \(2.04\,\mathrm{mol\,F_2} * \frac{38.00\,\mathrm{g}}{1\,\mathrm{mol\,F_2}} = 77.52\,\mathrm{g\,F_2}\)

Adjust the mass of F2 needed based on the reaction yield

Now we need to take into account the reaction yield which is 78.1%. This means that the actual mass of F2 consumed is greater than what we calculated in step 4. To find the mass of F2 required, we will divide the mass calculated in step 4 by the reaction yield: \(\frac{77.52\,\mathrm{g\,F_2}}{0.781} = 99.25\,\mathrm{g\,F_2}\) Therefore, 99.25 g of F2 is needed to produce 120 g of PF3 with a 78.1% reaction yield.

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, called reactants, transform into different substances known as products. These transformations involve breaking and forming of chemical bonds. In the context of the given exercise, we deal with a reaction between phosphorus (\(\mathrm{P}_4(s)\)) and fluorine (\(\mathrm{F}_2(g)\)), which produces phosphorus trifluoride (\(\mathrm{PF}_3(g)\)).
Understanding how these substances interact is key to mastering stoichiometry. The very first step in analyzing any chemical reaction is to balance the chemical equation. This involves ensuring that the number of each type of atom is equal on both sides of the equation.
For our reaction, if we start with the unbalanced equation:\[\mathrm{P}_4(s) + \mathrm{F}_2(g) \longrightarrow \mathrm{PF}_3(g)\]We will need to adjust the coefficients to this balanced form:\[\mathrm{P}_4(s) + 6\mathrm{F}_2(g) \longrightarrow 4\mathrm{PF}_3(g)\]This tells us that 1 molecule of \(\mathrm{P}_4\) reacts with 6 molecules of \(\mathrm{F}_2\) to form 4 molecules of \(\mathrm{PF}_3\).

Mole Concept

The mole concept is a fundamental aspect of chemistry that allows chemists to count particles like atoms, molecules, and ions in a given sample easily. This concept is based on Avogadro's number, which is \(6.022 \times 10^{23}\), the number of particles in one mole of a substance.
In our problem, we've been asked to find out how much \(\mathrm{F}_2\) is needed to create a specific amount of \(\mathrm{PF}_3\), which requires using the mole concept.
To do this, we first convert the given mass of \(\mathrm{PF}_3\) (120 g) to moles. The molar mass of \(\mathrm{PF}_3\) is needed for this conversion. Using periodic table values:

• Phosphorus (P): 30.97 g/mol
• Fluorine (F): 19.00 g/mol

The molar mass of \(\mathrm{PF}_3\) is \(30.97 + 3 \times 19.00 = 87.97\) g/mol.
Then, we convert the mass of \(\mathrm{PF}_3\) to moles:\[\frac{120 \, \mathrm{g}}{87.97 \, \mathrm{g/mol}} = 1.36 \, \mathrm{mol \ PF_3}\]This calculated amount in moles can help us find the moles of reactants needed.

Reaction Yield

Reaction yield is a measure of how much product is obtained from a chemical reaction relative to the theoretical maximum amount possible. In other words, it's a reflection of the efficiency of the reaction. Theoretical yield is the amount of product predicted by stoichiometry, while actual yield is what is actually obtained from performing the experiment.
For the exercise, the reaction yield is given as 78.1%, which indicates that only 78.1% of the theoretical yield is obtained in practice.
In calculating the mass of \(\mathrm{F}_2\) needed:

• The theoretical calculation, based on the stoichiometry, found that 77.52 g of \(\mathrm{F}_2\) would be needed.
• However, because of the 78.1% yield, more \(\mathrm{F}_2\) is necessary to produce the desired 120 g of \(\mathrm{PF}_3\).

To adjust for this, divide the mass calculated by the yield percentage:\[\frac{77.52 \, \mathrm{g \, F_2}}{0.781} = 99.25 \, \mathrm{g \, F_2}\]This means we need 99.25 g of \(\mathrm{F}_2\) to ensure the production of 120 g of \(\mathrm{PF}_3\) under the given conditions.