Answer:
Approximately [tex]60.3\; \rm g[/tex].
Explanation:
Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex] on a modern periodic table:
Calculate the formula mass of [tex]\rm AlBr_3[/tex] and [tex]\rm Br_2[/tex]:
[tex]\begin{aligned}& M(\mathrm{AlBr_3}) = 26.982 + 3 \times 79.904 \approx 266.694\; \rm g \cdot mol^{-1} \\ & M(\mathrm{Br_2}) = 2\times 79.904 \approx 159.808\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of moles of formula units in [tex]67.1\; \rm g[/tex] of [tex]\rm AlBr_3[/tex]:
[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{m(\mathrm{AlBr_3})}{M(\mathrm{AlBr_3})} \\ &\approx \frac{67.1\; \rm g}{266.694\; \rm g \cdot mol^{-1}} \approx 0.2516\; \rm mol \end{aligned}[/tex].
Refer to the balanced equation for this reaction. The ratio between the coefficients of [tex]\rm Br_2[/tex] and [tex]\rm AlBr_3[/tex] in that equation is three-to-two. That corresponds to the ratio:
[tex]\begin{aligned}\frac{n(\text{$\mathrm{Br_2}$, consumed})}{n(\text{$\mathrm{AlBr_3}$, produced})} &= \frac{3}{2}\end{aligned}[/tex].
It is already calculated that approximately [tex]0.2516\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] was produced through this reaction. Apply this ratio to approximate the (minimum) number of moles of [tex]\rm Br_2[/tex] that is consumed:
[tex]\displaystyle \frac{3}{2} \times 0.2516\; \rm mol \approx 0.3774\; \rm mol[/tex].
Calculate the mass of that [tex]0.3774\; \rm mol[/tex] of [tex]\rm Br_2[/tex]:
[tex]\begin{aligned}m(\mathrm{Br_2}) &= n(\mathrm{Br_2})\cdot M(\mathrm{Br_2}) \\ &\approx 0.3774\; \rm mol \times 159.808\; \rm g \cdot mol^{-1} \approx 60.3\; \rm g\end{aligned}[/tex].
