Stoicheometry question: Missing step?

Question

I've done this problem but it was wrong. I calculated my limiting reagent, how much Aluminum i need, calculated my theoretical yield. is there a step i'm missing? please show work so i see where i went wrong (at least for the missing step)

-Aluminum iodide is a reagent used in synthetic organic chemistry. It can be prepared by reacting aluminum metal with solid iodine. If 256 g aluminum metal is reacted with 3780 g iodine and the reaction proceeds in a 41.72% yield, what percent (by mass) of the remaining starting material is aluminum metal?

Answer 1

Hello Naomi

 

The balanced chemical reaction between Aluminum and Iodine is:

 

2 Al (s) + 3 I₂ (s) → 2 AlI₃ (s)

 

First let's determine the limiting reagent by converting mass of Al and I₂ to moles of AlI₃.  Molar masses for Al, I₂ and AlI₃ are 27, 254 and 408 g/mol, respectively.  From the reaction, you can see that 2 moles of Al reacts to form 2 moles of AlI₃, and 3 moles of I₂ reacts to form 2 moles of AlI₃.

 

256 g Al x (1 mol Al / 27 g Al) x (2 mol AlI₃ / 2 mol Al) = 9.48 mol AlI₃

 

3780 g I₂ x (1 mol I₂ / 254 g I₂) x (2 mol AlI₃ / 3 mol I₂) = 9.92 mol AlI₃

 

Since Al produces the lower amount of AlI₃, it will be our limiting reagent and the theoretical mass (yield) of AlI₃ formed is:

 

9.48 mol AlI₃ x (408 g AlI₃ / 1 mol AlI₃) = 3868 g AlI₃

 

From this reaction, the %-yield of AlI₃ is 41.72%.  %-Yield = (Actual yield / Theoretical yield) x 100%...  The actual yield of AlI₃ in grams is 41.72% x 3868 g = 1614 g AlI₃

 

From this mass (1614 g), let's convert it back to g of Al to determine how much Aluminum out of 256 g was consumed in the reaction.

 

1614 g AlI₃ x (1 mol AlI₃ / 408 g AlI₃) x (2 mol Al / 2 mol AlI₃) x (27 g Al / 1 mol Al) = 107 g Al

 

Mass of Al unreacted (remaining) = 256 - 107 = 149 g Al

 

%-Aluminum remaining = (149 / 256) x 100% = 58.20%