Final answer:
Approximately 445.2 grams of bromine (Br₂) are needed to react exactly with 50.1 grams of aluminum (Al).
Step-by-step explanation:
To calculate the number of grams of bromine (Br₂) needed to react with 50.1 g of aluminum (Al), we need to use the balanced chemical equation and stoichiometry.
The balanced equation for the reaction is: 2Al(s) + 3Br₂(l) -> 2AlBr₃(s).
From the stoichiometry of the equation, we can see that 2 moles of aluminum react with 3 moles of bromine. Therefore, to calculate the number of moles of bromine needed, we can set up a ratio based on the molar masses of aluminum and bromine.
In this case, we have 50.1 g of aluminum, which is equivalent to 50.1 g / 26.98 g/mol = 1.8578 mol of aluminum. From the balanced equation, we know that the ratio of moles of aluminum to moles of bromine is 2:3. So, we can set up the following ratio:
2 mol Al / 3 mol Br₂ = 1.8578 mol Al / x mol Br₂
Solving for x, we can find the number of moles of bromine needed:
x = (3 mol Br₂ * 1.8578 mol Al) / 2 mol Al = 2.7867 mol Br₂
Finally, to convert the number of moles of bromine to grams, we can use the molar mass of bromine (Br₂ = 159.808 g/mol):
Mass of bromine = 2.7867 mol Br₂ * 159.808 g/mol = 445.2 g Br₂
Therefore, approximately 445.2 grams of bromine (Br₂) are needed to react exactly with 50.1 grams of aluminum (Al).