Question

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for the reaction shown, compute the theoretical yield of the product in grams for each of
the following initial amounts of reactants
2 al(s) + 3 cl2(g) → 2 alcl3(s)
part b
correct
the theoretical yield is the amount of product made in a chemical reaction based on the amount of limiting reactant. for this chemical reaction, cl2 produced the least amount of product
5.5 g al, 19.8 g cl2
express your answer using three significant figures.
previous answers
2.18 g
part c
all attempts used; correct answer withheld by instructor
the theoretical yield is the amount of product made in a chemical reaction based on the amount of limiting reactant. for this chemical reaction, cl2 produced the least amount of product
0.439 g al, 2.29 g cl2
express your answer using three significant figures.

Explanation:

Step1: Calculate moles of reactants

• Molar mass of \(Al\): \(M_{Al}=26.98\ g/mol\). Moles of \(Al\), \(n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{0.439\ g}{26.98\ g/mol}\approx0.0163\ mol\)
• Molar mass of \(Cl_{2}\): \(M_{Cl_{2}} = 70.90\ g/mol\). Moles of \(Cl_{2}\), \(n_{Cl_{2}}=\frac{m_{Cl_{2}}}{M_{Cl_{2}}}=\frac{2.29\ g}{70.90\ g/mol}\approx0.0323\ mol\)

Step2: Determine limiting reactant

From the balanced equation \(2Al(s)+3Cl_{2}(g)\to2AlCl_{3}(s)\), the mole ratio of \(Al\) to \(Cl_{2}\) is \(\frac{n_{Al}}{n_{Cl_{2}}}=\frac{2}{3}\).
For \(n_{Al} = 0.0163\ mol\), required \(n_{Cl_{2}}\) (if \(Al\) is limiting) is \(n_{Cl_{2}}^{req}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0.0163\ mol = 0.02445\ mol\). Since \(0.0323\ mol>0.02445\ mol\), \(Al\) is the limiting reactant.

Step3: Calculate moles of \(AlCl_{3}\)

From the balanced equation, mole ratio of \(Al\) to \(AlCl_{3}\) is \(1:1\). So \(n_{AlCl_{3}}=n_{Al}=0.0163\ mol\)

Step4: Calculate mass of \(AlCl_{3}\)

Molar mass of \(AlCl_{3}\): \(M_{AlCl_{3}}=26.98+(3\times35.45)=133.33\ g/mol\)
Mass of \(AlCl_{3}\), \(m_{AlCl_{3}}=n_{AlCl_{3}}\times M_{AlCl_{3}}=0.0163\ mol\times133.33\ g/mol\approx2.17\ g\)

Answer:

\(2.17\ g\)