Answer :
Sure, let's break it down step by step with all the needed explanations and calculations.
1. Determine the molar mass of [tex]\( AlCl_3 \)[/tex]:
- The molar mass of Aluminum (Al) = 26.98 g/mol
- The molar mass of Chlorine (Cl) = 35.45 g/mol
- For [tex]\( AlCl_3 \)[/tex], the molar mass = 26.98 + 3 * 35.45 = 133.33 g/mol
2. Calculate the number of moles of [tex]\( AlCl_3 \)[/tex]:
- Given mass of [tex]\( AlCl_3 \)[/tex] = 362 grams
- Moles of [tex]\( AlCl_3 \)[/tex] = [tex]\( \frac{362}{133.33} \approx 2.715067876696917 \)[/tex] moles
3. Use stoichiometry to determine the moles of [tex]\( Br_2 \)[/tex] required:
- According to the balanced equation:
[tex]\[ 2 AlCl_3 + 3 Br_2 \rightarrow 2 AlBr_3 + 3 Cl_2 \][/tex]
- The mole ratio of [tex]\( AlCl_3 \)[/tex] to [tex]\( Br_2 \)[/tex] is 2:3. This means that for every 2 moles of [tex]\( AlCl_3 \)[/tex], 3 moles of [tex]\( Br_2 \)[/tex] are required.
- Thus, moles of [tex]\( Br_2 \)[/tex] required = [tex]\( \frac{3}{2} \times 2.715067876696917 \approx 4.072601815045376 \)[/tex] moles
4. Calculate the mass of [tex]\( Br_2 \)[/tex] required:
- The molar mass of Bromine (Br) = 79.90 g/mol
- Because [tex]\( Br_2 \)[/tex] is diatomic, the molar mass of [tex]\( Br_2 \)[/tex] = 2 79.90 = 159.8 g/mol
- Therefore, mass of [tex]\( Br_2 \)[/tex] required = moles of [tex]\( Br_2 \)[/tex] molar mass of [tex]\( Br_2 \)[/tex]
- Mass of [tex]\( Br_2 \)[/tex] = 4.072601815045376 moles * 159.8 g/mol = 650.801770044251 grams
Thus, 650.801770044251 grams of [tex]\( Br_2 \)[/tex] are required to react completely with 362 grams of [tex]\( AlCl_3 \)[/tex]. This matches none of the options given exactly, but from the context, assuming there might be a typographical error, it is safe to say that 765 grams is the closest provided option if rounded differently, or none of the provided options is entirely correct considering precision.
1. Determine the molar mass of [tex]\( AlCl_3 \)[/tex]:
- The molar mass of Aluminum (Al) = 26.98 g/mol
- The molar mass of Chlorine (Cl) = 35.45 g/mol
- For [tex]\( AlCl_3 \)[/tex], the molar mass = 26.98 + 3 * 35.45 = 133.33 g/mol
2. Calculate the number of moles of [tex]\( AlCl_3 \)[/tex]:
- Given mass of [tex]\( AlCl_3 \)[/tex] = 362 grams
- Moles of [tex]\( AlCl_3 \)[/tex] = [tex]\( \frac{362}{133.33} \approx 2.715067876696917 \)[/tex] moles
3. Use stoichiometry to determine the moles of [tex]\( Br_2 \)[/tex] required:
- According to the balanced equation:
[tex]\[ 2 AlCl_3 + 3 Br_2 \rightarrow 2 AlBr_3 + 3 Cl_2 \][/tex]
- The mole ratio of [tex]\( AlCl_3 \)[/tex] to [tex]\( Br_2 \)[/tex] is 2:3. This means that for every 2 moles of [tex]\( AlCl_3 \)[/tex], 3 moles of [tex]\( Br_2 \)[/tex] are required.
- Thus, moles of [tex]\( Br_2 \)[/tex] required = [tex]\( \frac{3}{2} \times 2.715067876696917 \approx 4.072601815045376 \)[/tex] moles
4. Calculate the mass of [tex]\( Br_2 \)[/tex] required:
- The molar mass of Bromine (Br) = 79.90 g/mol
- Because [tex]\( Br_2 \)[/tex] is diatomic, the molar mass of [tex]\( Br_2 \)[/tex] = 2 79.90 = 159.8 g/mol
- Therefore, mass of [tex]\( Br_2 \)[/tex] required = moles of [tex]\( Br_2 \)[/tex] molar mass of [tex]\( Br_2 \)[/tex]
- Mass of [tex]\( Br_2 \)[/tex] = 4.072601815045376 moles * 159.8 g/mol = 650.801770044251 grams
Thus, 650.801770044251 grams of [tex]\( Br_2 \)[/tex] are required to react completely with 362 grams of [tex]\( AlCl_3 \)[/tex]. This matches none of the options given exactly, but from the context, assuming there might be a typographical error, it is safe to say that 765 grams is the closest provided option if rounded differently, or none of the provided options is entirely correct considering precision.
