Question

consider the following unbalanced reaction:

p_{4}(s)+f_{2}(g)
ightarrow pf_{3}(g)

what mass of ( f_{2} ) is needed to produce 100. g of ( pf_{3} ) if the reaction has a ( 73.1 % ) yield?

mass ( = ) g

Explanation:

Step1: Balance the chemical equation

$$\mathrm{P_4(s)+6F_2(g)\to 4PF_3(g)}$$

Step2: Calculate the moles of $\mathrm{PF_3}$ produced (actual yield)

The molar mass of $\mathrm{PF_3}$ is $M = 30.97+(3\times19.00)=87.97\ \mathrm{g/mol}$.
The moles of $\mathrm{PF_3}$ (actual) $n_{actual}=\frac{m}{M}=\frac{100\ \mathrm{g}}{87.97\ \mathrm{g/mol}}\approx1.137\ \mathrm{mol}$

Step3: Calculate the moles of $\mathrm{PF_3}$ (theoretical yield)

Using the formula $\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$.
Let $n_{theoretical}$ be the moles of $\mathrm{PF_3}$ (theoretical). Then $n_{theoretical}=\frac{n_{actual}}{\text{Percent yield}/100}=\frac{1.137\ \mathrm{mol}}{0.731}\approx1.555\ \mathrm{mol}$

Step4: Use mole - ratio from balanced equation to find moles of $\mathrm{F_2}$

From the balanced equation $\mathrm{P_4 + 6F_2\to4PF_3}$, the mole ratio of $\mathrm{F_2}$ to $\mathrm{PF_3}$ is $\frac{6}{4}$.
So moles of $\mathrm{F_2}, n_{F_2}=\frac{6}{4}\times n_{theoretical}=\frac{6}{4}\times1.555\ \mathrm{mol}=2.3325\ \mathrm{mol}$

Step5: Calculate the mass of $\mathrm{F_2}$

The molar mass of $\mathrm{F_2}$ is $M = 2\times19.00 = 38.00\ \mathrm{g/mol}$.
Mass of $\mathrm{F_2}, m = n\times M=2.3325\ \mathrm{mol}\times38.00\ \mathrm{g/mol}\approx88.6\ \mathrm{g}$

Answer:

$88.6$