Question

consider the following unbalanced reaction:
p4(s) + f2(g) → pf3(g)
what mass of f2 is needed to produce 100. g of pf3 if the reaction has a 78.1% yield?
mass = 82.9 g
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Explanation:

Step1: Write the balanced chemical equation

The balanced chemical equation for the reaction is \(P_4 + 6F_2
ightarrow4PF_3\).

Step2: Calculate the theoretical yield

The molar mass of \(PF_3\) is \(M_{PF_3}=30.97+(3\times19.00)=87.97\ g/mol\).
If the actual yield of \(PF_3\) is \(m = 100\ g\) and the percent - yield \(y = 78.1\%\), we use the formula \(y=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%\).
Let the theoretical yield be \(m_{theo}\). Then \(m_{theo}=\frac{m}{y}\times100=\frac{100\ g}{78.1\%}\times100 = 128.04\ g\).

Step3: Calculate the moles of \(PF_3\) (theoretical)

The number of moles of \(PF_3\), \(n_{PF_3}=\frac{m_{theo}}{M_{PF_3}}=\frac{128.04\ g}{87.97\ g/mol}=1.455\ mol\).

Step4: Calculate the moles of \(F_2\)

From the balanced equation \(P_4 + 6F_2
ightarrow4PF_3\), the mole ratio of \(F_2\) to \(PF_3\) is \(\frac{n_{F_2}}{n_{PF_3}}=\frac{6}{4}\).
So \(n_{F_2}=\frac{6}{4}\times n_{PF_3}=\frac{6}{4}\times1.455\ mol = 2.1825\ mol\).

Step5: Calculate the mass of \(F_2\)

The molar mass of \(F_2\) is \(M_{F_2}=2\times19.00 = 38.00\ g/mol\).
The mass of \(F_2\), \(m_{F_2}=n_{F_2}\times M_{F_2}=2.1825\ mol\times38.00\ g/mol = 82.9\ g\).

Answer:

\(82.9\ g\)