Answer:
Approximately [tex]1.29 \times 10^3[/tex] grams.
Explanation:
Let [tex]x[/tex] represent the number of grams of aluminum iodide required to yield that 73.75 grams of aluminum.
In most cases, the charge on each aluminum ion would be +3 while the charge on each iodide ion would be -1. For the charges to balance, there needs to be three iodide ions for every aluminum ion. Hence, the empirical formula for aluminum iodide would be [tex]\rm AlI_3[/tex].
How many moles of formula units in that [tex]x[/tex] grams of [tex]\rm AlI_3[/tex]? Start by calculating its formula mass [tex]M(\mathrm{AlI_3})[/tex]. Look up the relative atomic mass of aluminum and iodine on a modern periodic table:
[tex]M(\mathrm{AlI_3}) = 1\times 26.982 + 3\times 126.904 = 410.694\; \rm g \cdot mol^{-1}[/tex].
[tex]n(\mathrm{AlI_3}) = \displaystyle \frac{m}{M} = \frac{x}{410.694}\;\rm mol[/tex].
Since there's one aluminum ion in every formula unit,
[tex]n(\mathrm{Al}) = n(\mathrm{AlI_3}) = \displaystyle \frac{x}{410.694}\; \rm mol[/tex].
How many grams of aluminum would that be?
[tex]m(\mathrm{Al}) = n \cdot M = \displaystyle \frac{x}{410.694}\; \times 26.982 = \frac{26.982}{410.694}\, x\; \rm g[/tex].
However, since according to the question, the percentage yield (of aluminum) is only [tex]86.8\%[/tex]. Hence, the actual yield of aluminum would be:
[tex]\begin{aligned}&\text{Actual Yield} \\ &= \text{Percentage Yield} \times \text{Theoretical Yield} \\ &= 86.8\% \times \frac{26.982}{410.694}\, x \\ &= 0.868 \times \frac{26.982}{410.694}\, x \\ &\approx 0.0570263\, x\; \rm g\end{aligned}[/tex].
Given that the actual yield is 73.75 grams,
[tex]0.0570263\, x = 73.75[/tex].
[tex]\displaystyle x = \frac{73.75}{0.0570263} \approx 1.29 \times 10^3\; \rm g[/tex].
