Answer: 181.16 grams of bromine (Br2) will react completely with 20.4 grams of aluminum (Al) to form aluminum bromide (AlBr3).
Step-by-step explanation:
To calculate the mass of bromine (Br2) that will react completely with 20.4 g of aluminum (Al) to form aluminum bromide (AlBr3), we need to determine the balanced chemical equation for the reaction and use stoichiometry.
1. Balanced chemical equation:
The balanced equation for the reaction between aluminum and bromine to form aluminum bromide is:
2 Al + 3 Br2 → 2 AlBr3
2. Calculate the molar mass:
The molar mass of bromine (Br2) is 159.808 g/mol.
3. Convert the mass of aluminum (Al) to moles:
Using the molar mass of aluminum (Al) (26.98 g/mol), we can convert 20.4 g of Al to moles:
20.4 g Al * (1 mol Al / 26.98 g Al) = 0.756 mol Al
4. Use stoichiometry to determine the moles of bromine (Br2):
From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of bromine. Therefore, the moles of bromine can be calculated as:
0.756 mol Al * (3 mol Br2 / 2 mol Al) = 1.134 mol Br2
5. Convert moles of bromine (Br2) to grams:
Using the molar mass of bromine (Br2) (159.808 g/mol), we can convert moles of Br2 to grams:
1.134 mol Br2 * (159.808 g Br2 / 1 mol Br2) = 181.16 g Br2
Therefore, 181.16 grams of bromine (Br2) will react completely with 20.4 grams of aluminum (Al) to form aluminum bromide (AlBr3).
I hope this helps :)