Identifying Lewis Acids: Examples and Corrections

[vsage]

It's been awhile since I've done chem and I don't really remember how to identify Lewis Acids too well so here are a few problems I'd like you guys to check over / correct:

$$CaO(s) + SO_3(g) \rightarrow CaSO_4$$ My answer: $$SO_3$$

$$Ca^{+2}(aq) + SO_4^{-2}(aq) \rightarrow CaSO_4(s)$$ My answer: $$Ca^{+2}$$

$$Al(OH_2)_6^{+3} + H_2O \rightarrow Al(OH_2)_5OH^{+2} + H_3O^{+}$$ My answer: $$Al(OH_2)_5OH^{+2}$$

$$CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(s)$$. My answer: $$CaO(s)$$ but there's also another Lewis Acid according to the worksheet

$$NaCl(s) \rightarrow (H_2O) Na^+(aq) + Cl^-(aq)$$
My answer: $$Cl^-$$ but there's another Lewis Acid according to the worksheet

I unfortunately had to miss class for another exam the day of this lecture so I'm pretty clueless! Thanks in advance.

 

[wais]

First of all, your answers for the first three problems are correct. SO_3, Ca^{+2}, and Al(OH_2)_5OH^{+2} are all Lewis acids in these reactions.

However, for the fourth problem, your answer is not completely correct. While CaO can act as a Lewis acid in this reaction, there is actually another Lewis acid present as well - H_2O. In this reaction, H_2O acts as a Lewis acid by accepting a lone pair of electrons from CaO to form Ca(OH)_2. So the correct answer would be CaO and H_2O as Lewis acids in this reaction.

For the final problem, your answer is partially correct. Cl^- is indeed a Lewis acid in this reaction as it accepts a lone pair of electrons from H_2O to form HCl. However, there is another Lewis acid present - Na^+. Na^+ acts as a Lewis acid by accepting a lone pair of electrons from H_2O to form NaOH. So the correct answer would be Cl^- and Na^+ as Lewis acids in this reaction.

In summary, it is important to remember that Lewis acids are electron acceptors and can be identified by their ability to accept a lone pair of electrons from a Lewis base. In some cases, there may be multiple Lewis acids present in a reaction. Keep practicing and you will become more confident in identifying Lewis acids!