Question

use the following information to answer the next two questions.
the formation of aluminum iodide is shown in the unbalanced reaction below

1. if 1.20 g of aluminum and 6.00 g of iodine are allowed to react, the mass of aluminum iodide formed will be______g.

(record your answer in the numerical response section below.)
your answer:

1. if 4.5 g of aluminum iodide are actually formed, the percent yield of the reaction is______%.

(record your answer in the numerical response section below.)
your answer:

Explanation:

Step1: Balance the chemical equation

$$2Al_{(s)} + 3I_{2(s)} ightarrow 2AlI_{3(s)}$$

Step2: Calculate the molar masses

The molar mass of \(Al\) is \(M_{Al}=26.98\ g/mol\), the molar mass of \(I_{2}\) is \(M_{I_{2}} = 2\times126.90=253.80\ g/mol\), and the molar mass of \(AlI_{3}\) is \(M_{AlI_{3}}=26.98 + 3\times126.90=407.68\ g/mol\)

Step3: Calculate the moles of reactants

Moles of \(Al\), \(n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{1.20\ g}{26.98\ g/mol}\approx0.0445\ mol\)
Moles of \(I_{2}\), \(n_{I_{2}}=\frac{m_{I_{2}}}{M_{I_{2}}}=\frac{6.00\ g}{253.80\ g/mol}\approx0.0236\ mol\)

Step4: Determine the limiting reactant

From the balanced equation, the mole ratio of \(Al\) to \(I_{2}\) is \(2:3\).
For \(n_{Al} = 0.0445\ mol\), the moles of \(I_{2}\) required is \(n_{I_{2}\text{(required)}}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0.0445 = 0.0668\ mol\)
Since \(n_{I_{2}\text{(available)}}=0.0236\ mol<0.0668\ mol\), \(I_{2}\) is the limiting reactant.

Step5: Calculate the moles of \(AlI_{3}\) formed

From the balanced equation, \(n_{AlI_{3}}=\frac{2}{3}n_{I_{2}}\)
\(n_{AlI_{3}}=\frac{2}{3}\times0.0236\ mol\approx0.0157\ mol\)

Step6: Calculate the mass of \(AlI_{3}\) formed

\(m_{AlI_{3}}=n_{AlI_{3}}\times M_{AlI_{3}}\)
\(m_{AlI_{3}}=0.0157\ mol\times407.68\ g/mol\approx6.40\ g\)

Step7: Calculate the percent - yield (for question 15)

Percent yield \(=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%\)
Given actual yield \(m_{actual}=4.5\ g\) and theoretical yield \(m_{theoretical} = 6.40\ g\)
Percent yield \(=\frac{4.5\ g}{6.40\ g}\times100\% \approx 70.3\%\)

Answer:

1. \(6.40\)
2. \(70.3\)