Answer :
Approximately 43.32 grams of [tex]Br_2[/tex] are required to react completely with 36.2 grams of [tex]AlCl_3[/tex].
To solve this, we need to find the molar mass of [tex]AlCl_3[/tex] and [tex]Br_2[/tex]. Then, we can use stoichiometry to determine the amount of [tex]Br_2[/tex] required.
The molar mass of [tex]AlCl_3[/tex] is 26.98 + (3 * 35.45) = 133.33 g/mol.
The molar mass of [tex]Br_2[/tex] is 2 * 79.90 = 159.80 g/mol.
The balanced equation shows a 1:1 molar ratio between [tex]AlCl_3[/tex] and [tex]Br_2[/tex].
So, we can set up a proportion to find out how many grams of [tex]Br_2[/tex] are needed:
(36.2 g [tex]AlCl_3[/tex] ) / (133.33 g/mol [tex]AlCl_3[/tex]) = (x g [tex]Br_2[/tex]) / (159.80 g/mol [tex]Br_2[/tex])
Solving for x, we get:
x = (36.2 g * 159.80 g/mol) / 133.33 g/mol ≈ 43.32 g
So, approximately 43.32 grams of [tex]Br_2[/tex] are required to react completely with 36.2 grams of [tex]AlCl_3[/tex].
Complete question :
Based on the equation, how many grams of [tex]Br_2[/tex] are required to react completely with 36.2 grams of [tex]AlCl_3[/tex]?
[tex]AlCl_3 + Br_2 -- > AlBr_3 + Cl_2[/tex]
