Question

question 17
what is the maximum amount, in g, of p₂i₄ that can be prepared from 8.75g of p₄o₆ and 11.08 g of iodine, according to the following reaction:
5 p₄o₆ + 8 i₂ → 4 p₂i₄ + 3 p₄o₁₀
report your answer to 3 significant figures.

Explanation:

Step1: Calculate moles of \(P_4O_6\)

The molar mass of \(P_4O_6\) is \(M_{P_4O_6}=(4\times30.97)+(6\times16)=219.88\space g/mol\).
Using the formula \(n = \frac{m}{M}\), for \(P_4O_6\), \(n_{P_4O_6}=\frac{8.75\space g}{219.88\space g/mol}\approx0.0398\space mol\)

Step2: Calculate moles of \(I_2\)

The molar mass of \(I_2\) is \(M_{I_2}=2\times126.90 = 253.8\space g/mol\).
Using the formula \(n=\frac{m}{M}\), for \(I_2\), \(n_{I_2}=\frac{11.08\space g}{253.8\space g/mol}\approx0.0436\space mol\)

Step3: Determine the limiting reactant

From the balanced equation \(5P_4O_6 + 8I_2
ightarrow4P_2I_4+3P_4O_{10}\), the mole ratio of \(P_4O_6\) to \(I_2\) is \(\frac{n_{P_4O_6}}{n_{I_2}}=\frac{5}{8}\).
If we assume \(P_4O_6\) is the limiting reactant, the required \(n_{I_2}\) for \(n_{P_4O_6} = 0.0398\space mol\) is \(n_{I_2}^{required}=0.0398\times\frac{8}{5}=0.0637\space mol\). But we have \(n_{I_2} = 0.0436\space mol\).
If we assume \(I_2\) is the limiting reactant, the required \(n_{P_4O_6}\) for \(n_{I_2}=0.0436\space mol\) is \(n_{P_4O_6}^{required}=0.0436\times\frac{5}{8}=0.0273\space mol\). Since \(0.0273\space mol<0.0398\space mol\), \(I_2\) is the limiting reactant.

Step4: Calculate moles of \(P_2I_4\)

From the balanced equation, the mole ratio of \(I_2\) to \(P_2I_4\) is \(\frac{n_{I_2}}{n_{P_2I_4}}=\frac{8}{4} = 2\).
Since \(n_{I_2}=0.0436\space mol\), then \(n_{P_2I_4}=\frac{0.0436\space mol}{2}=0.0218\space mol\)

Step5: Calculate mass of \(P_2I_4\)

The molar mass of \(P_2I_4\) is \(M_{P_2I_4}=(2\times30.97)+(4\times126.90)=569.54\space g/mol\)
Using \(m = n\times M\), \(m_{P_2I_4}=0.0218\space mol\times569.54\space g/mol\approx12.4\space g\)

Answer:

\(12.4\)