You'll need to provide the balanced chemical equation for the reaction to calculate the grams of aluminum bromide formed.

Here's how to solve the problem once you have the equation:

1. Balanced Chemical Equation:

* The equation you provide will show the stoichiometric ratio between HBr and aluminum bromide (AlBr₃).

2. Molar Masses:

* Determine the molar mass of HBr (1 g/mol H + 79.9 g/mol Br = 80.9 g/mol)

* Determine the molar mass of AlBr₃ (26.98 g/mol Al + 3 * 79.9 g/mol Br = 266.7 g/mol)

3. Convert Grams of HBr to Moles:

* Divide the given mass of HBr (121 g) by its molar mass (80.9 g/mol):

121 g HBr / 80.9 g/mol = 1.50 mol HBr

4. Use Stoichiometry to Find Moles of AlBr₃:

* Use the mole ratio from the balanced equation to determine the moles of AlBr₃ produced from 1.50 mol HBr.

* For example, if the balanced equation shows a 3:1 mole ratio between HBr and AlBr₃, then:

1.50 mol HBr * (1 mol AlBr₃ / 3 mol HBr) = 0.50 mol AlBr₃

5. Convert Moles of AlBr₃ to Grams:

* Multiply the moles of AlBr₃ (0.50 mol) by its molar mass (266.7 g/mol):

0.50 mol AlBr₃ * 266.7 g/mol = 133.35 g AlBr₃

Therefore, you would need to provide the balanced chemical equation for the reaction to determine the exact grams of aluminum bromide formed.