Sure, let's solve this step-by-step to determine how many grams of bromine ([tex]\(Br_2\)[/tex]) are required to react completely with 29.2 grams of aluminum chloride ([tex]\(AlCl_3\)[/tex]) based on the balanced chemical equation:
[tex]\[ 2 \, AlCl_3 + 3 \, Br_2 \rightarrow 2 \, AlBr_3 + 3 \, Cl_2 \][/tex]
1. Calculate the molar mass of [tex]\(\text{AlCl}_3\)[/tex].
The molar mass can be found by adding together the atomic masses of aluminum (Al) and chlorine (Cl):
[tex]\[ \text{Molar mass of } Al = 26.98 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } Cl = 35.45 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } AlCl_3 = 26.98 + 3 \times 35.45 = 133.33 \, \text{g/mol} \][/tex]
2. Calculate the number of moles of [tex]\(\text{AlCl}_3\)[/tex] in 29.2 grams.
Using the molar mass calculated:
[tex]\[ \text{Moles of } AlCl_3 = \frac{29.2 \, \text{grams}}{133.33 \, \text{g/mol}} = 0.219 \text{ moles} \][/tex]
3. Determine the moles of [tex]\(\text{Br}_2\)[/tex] required.
According to the balanced equation, 2 moles of [tex]\(\text{AlCl}_3\)[/tex] react with 3 moles of [tex]\(\text{Br}_2\)[/tex]:
[tex]\[ \text{Moles of } Br_2 = \left(\frac{3}{2}\right) \times \text{moles of } AlCl_3 = \left(\frac{3}{2}\right) \times 0.219 \, \text{moles} = 0.329 \, \text{moles} \][/tex]
4. Calculate the molar mass of [tex]\(\text{Br}_2\)[/tex].
[tex]\[ \text{Molar mass of } Br = 79.904 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } Br_2 = 2 \times 79.904 = 159.808 \, \text{g/mol} \][/tex]
5. Calculate the mass of [tex]\(\text{Br}_2\)[/tex] required.
Using the number of moles and the molar mass of [tex]\(\text{Br}_2\)[/tex]:
[tex]\[ \text{Mass of Br}_2 = 0.329 \, \text{moles} \times 159.808 \, \text{g/mol} = 52.498 \, \text{grams} \][/tex]
Therefore, the mass of [tex]\(Br_2\)[/tex] required to react completely with 29.2 grams of [tex]\(AlCl_3\)[/tex] is approximately 52.5 grams. The correct answer to the question is:
[tex]\[
\boxed{52.6 \text{ grams}}
\][/tex]
