Question

Given the reaction has a percent yield of 86.8 how many grams of aluminum iodide would be required to yield an actual amount of 73.75 grams of aluminum?

Answer 1

Answer:

Approximately $1.29 \times 10^3$ grams.

Explanation:

Let $x$ 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 $\rm AlI_3$.

How many moles of formula units in that $x$ grams of $\rm AlI_3$? Start by calculating its formula mass $M(\mathrm{AlI_3})$. Look up the relative atomic mass of aluminum and iodine on a modern periodic table:

• Al: 26.982.
• I: 126.904.

$M(\mathrm{AlI_3}) = 1\times 26.982 + 3\times 126.904 = 410.694\; \rm g \cdot mol^{-1}$.

$n(\mathrm{AlI_3}) = \displaystyle \frac{m}{M} = \frac{x}{410.694}\;\rm mol$.

Since there's one aluminum ion in every formula unit,

$n(\mathrm{Al}) = n(\mathrm{AlI_3}) = \displaystyle \frac{x}{410.694}\; \rm mol$.

How many grams of aluminum would that be?

$m(\mathrm{Al}) = n \cdot M = \displaystyle \frac{x}{410.694}\; \times 26.982 = \frac{26.982}{410.694}\, x\; \rm g$.

However, since according to the question, the percentage yield (of aluminum) is only $86.8\%$. Hence, the actual yield of aluminum would be:

$\begin{aligned}&\text{Actual Yield} \\ &= \text{Percentage Yield} \times \text{Theoretical Yield} \\ &= 86.8\% \times \frac{26.982}{410.694}\, x \\ &= 0.868 \times \frac{26.982}{410.694}\, x \\ &\approx 0.0570263\, x\; \rm g\end{aligned}$.

Given that the actual yield is 73.75 grams,

$0.0570263\, x = 73.75$.

$\displaystyle x = \frac{73.75}{0.0570263} \approx 1.29 \times 10^3\; \rm g$.