Answer:
239.6 mg is the theoretical yield of tribromoanisole.
Explanation:
[tex]C_6H_5-OCH_3+3Br_2\rightarrow C_6H_2Br_3-OCH_3+3HBr[/tex]
Mass of anisole [tex]C_6H_5-OCH_3[/tex] = 75 mg = 0.075 g
1 mg = 0.001 g
Moles of anisole = [tex]\frac{0.075 g}{108 g/mol}=0.0006944 mol[/tex]
According to reaction, 1 mole anisole gives 1 mole of tribromoanisole.
Then 0.0006944 mole of anisole will give:
[tex]\frac{1}{1}\times 0.0006944 mol=0.0006944 mol[/tex] of tribromoanisole
Mass of 0.0006944 mole of tribromoanisole:
0.0006944 mol × 345 g/mol = 0.2396 g = 239.6 mg
239.6 mg is the theoretical yield of tribromoanisole.
