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Chapter 3: Problem 116

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

Expert verified
To produce 120 g of PF₃ with a 78.1% yield, approximately 99.18 g of F₂ gas is needed.

Step by step solution

01

Balance the chemical equation

The unbalanced equation is given as: \(\mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g)\) To balance the equation, count the atoms of each element present on both sides of the equation. We have 4 P atoms and 2 F atoms in the reactants, while we have 1 P atom and 3 F atoms in the product. To balance it, distribute the coefficients: \[\mathrm{P}_{4}(s)+6\mathrm{F}_{2}(g) \longrightarrow 4\mathrm{PF}_{3}(g)\] Now, 4 P atoms and 12 F atoms are on both sides, so the equation is balanced.
02

Convert mass of the desired product to moles

We're given that 120 g of PF₃ is to be produced. To find out how much F₂ is needed, first convert the mass of PF₃ to moles using its molar mass: \[\text{Moles of PF}_{3} = \frac{\text{mass of PF}_{3}}{\text{molar mass of PF}_{3}}\] The molar mass of PF₃ is approximately 88 g/mol (1 P: 31 g/mol and 3 F: 57 g/mol), so: \[\text{Moles of PF}_{3} = \frac{120\,\text{g}}{88\,\text{g/mol}} = 1.36\,\text{mol}\]
03

Use stoichiometry to find moles of F₂ needed

Now we'll use the balanced chemical equation to relate the moles of PF₃ to the moles of F₂ needed. According to the balanced equation, 6 moles of F₂ are required to produce 4 moles of PF₃. So, we can apply this ratio: \[\text{Moles of F}_{2} = \text{Moles of PF}_{3} \times \frac{6\,\text{moles of F}_{2}}{4\,\text{moles of PF}_{3}}\] Substitute the moles of PF₃ calculated in Step 2: \[\text{Moles of F}_{2} = 1.36\,\text{mol}\times \frac{6}{4} = 2.04\,\text{mol}\]
04

Adjust for the percentage yield

The reaction has a 78.1% yield, so we'll need to account for that when determining the amount of F₂ needed. Since yield is expressed as actual yield divided by the theoretical yield, we can find the actual amount needed as follows: \[\text{Actual moles of F}_{2} = \frac{\text{Theoretical moles of F}_{2}}{\text{yield}} = \frac{2.04\,\text{mol}}{0.781} = 2.61\,\text{mol}\]
05

Convert moles of F₂ to mass

Finally, use the molar mass of F₂ to find the mass needed: \[\text{Mass of F}_{2} = \text{moles of F}_{2} \times \text{molar mass of F}_{2}\] The molar mass of F₂ is approximately 38 g/mol (2 F: 38 g/mol), so: \[\text{Mass of F}_{2} = 2.61\,\text{mol}\times 38\,\text{g/mol} \approx 99.18\,\text{g}\] So, about 99.18 g of F₂ gas is needed to produce 120 g of PF₃ with a 78.1% yield.

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