Final answer:
The mass of aluminum bromide that can be produced from 5.6 mol of Al and 4.4 mol of Br₂ is 781.21 grams, with bromine being the limiting reagent.
Step-by-step explanation:
The question requires the determination of the mass of aluminum bromide that can be produced when 5.6 mol of aluminum reacts with 4.4 mol of bromine, considering bromine as the limiting reagent. According to the balanced chemical equation 2 Al + 3 Br₂ → 2 AlBr₃, 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of aluminum bromide. Given that we have 5.6 moles of aluminum and 4.4 moles of bromine, we need to first find out which reactant is limiting. It can be observed that bromine will be the limiting reagent as less than the required stoichiometric amount is present for the reaction with aluminum.
Next, we calculate the amount of aluminum bromide that can be formed using the moles of limiting reagent (bromine):
- Molar ratio of Br₂ to AlBr₃ from the equation is 3:2.
- The moles of AlBr₃ that can be formed from 4.4 moles of Br₂ are (4.4 moles Br₂) × (2 moles AlBr₃ / 3 moles Br₂) = 2.93 moles AlBr₃.
Now we convert moles of aluminum bromide to mass:
- The molar mass of AlBr₃ is approximately 266.69 g/mol.
- The mass of AlBr₃ that can be produced is (2.93 moles AlBr₃) × (266.69 g/mol) = 781.21 grams.
Therefore, 781.21 grams of aluminum bromide can be produced when 5.6 mol of aluminum reacts with 4.4 mol of bromine, considering bromine as the limiting reagent.