Final answer:
To find the mass of AlBr₃ produced, convert 20.0 g of Al to moles, use the stoichiometry of the equation 2Al + 3Br₂ → 2AlBr₃ to find moles of AlBr₃, and multiply by its molar mass. The calculation shows that 197.6 g of AlBr₃ is produced.
Step-by-step explanation:
To determine the mass of AlBr₃ produced when 20.0 g of Al reacts completely with 30.0 g of Br₂, we need to convert the given masses of the reactants to moles using their molar masses (Al: 26.98 g/mol, Br₂: 159.808 g/mol). According to the balanced chemical equation 2Al + 3Br₂ → 2AlBr₃, the mole ratio of Al to AlBr₃ is 1:1, and Br₂ to AlBr₃ is 3:2. Since bromine is in excess, the reaction will be limited by the amount of aluminum. The molar mass of AlBr₃ is approximately 266.69 g/mol. Multiplying the moles of Al by the molar mass of AlBr₃ gives the mass of the product formed.
First, we calculate the moles of aluminum:
- moles of Al = mass (g) / molar mass (g/mol) = 20.0 g / 26.98 g/mol = 0.741 moles of Al
Using the stoichiometry of the reaction:
- moles of AlBr₃ = moles of Al = 0.741 moles
Then, calculate the mass of AlBr₃ produced:
- mass of AlBr₃ = moles of AlBr₃ x molar mass of AlBr₃ = 0.741 moles x 266.69 g/mol = 197.6 g of AlBr₃