Question

what quantity, in moles, of al are necessary to form 69.2 g of albr₃ from the following reaction?
2 al(s) + 3 br₂(l) ⇌ 2 albr₃(s)

Explanation:

Step1: Calculate the molar mass of \(AlBr_3\)

The molar mass of \(AlBr_3\) is \(M = 26.98\ g/mol+(3\times79.90\ g/mol)=266.68\ g/mol\).

Step2: Calculate the moles of \(AlBr_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 69.2\ g\) and \(M = 266.68\ g/mol\). So \(n_{AlBr_3}=\frac{69.2\ g}{266.68\ g/mol}\approx0.26\ mol\).

Step3: Use the stoichiometry of the reaction

From the balanced equation \(2Al(s)+3Br_2(l)
ightleftharpoons2AlBr_3(s)\), the mole ratio of \(Al\) to \(AlBr_3\) is \(2:2 = 1:1\). So \(n_{Al}=n_{AlBr_3}\).

Answer:

\(0.26\ mol\)