Question

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 a
1.0 g al, 1.0 g cl₂
express your answer using two significant figures.
1.3 g
part b
5.5 g al, 19.8 g cl₂
express your answer using three significant figures.
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.
part c
0.439 g al, 2.29 g cl₂
express your answer using three significant figures.

Explanation:

Step1: Calculate moles of reactants

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

Step2: Determine the limiting reactant

From the balanced equation \(2Al(s)+3Cl_{2}(g)\to2AlCl_{3}(s)\), the mole ratio of \(Al\) to \(Cl_{2}\) is \(2:3\).
For \(n_{Al} = 0.0163\ mol\), moles of \(Cl_{2}\) required \(n_{Cl_{2},req}=\frac{3}{2}\times0.0163\ mol = 0.0245\ mol\). Since \(n_{Cl_{2}}=0.0323\ mol>0.0245\ mol\), \(Al\) is the limiting reactant.

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

From the balanced equation, mole ratio of \(Al\) to \(AlCl_{3}\) is \(2:2 = 1:1\). So moles of \(AlCl_{3}\), \(n_{AlCl_{3}}=n_{Al}=0.0163\ mol\)

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:

\(2.17\ g\)