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I came across three reactions while studying p-block compounds in inorganic chemistry.

$$\ce{2Al + 2NaOH + 6H2O -> 2 Na[Al(OH)4] + 3H2} \label{eq:1} \tag{1}$$

$$\ce{Al2O3 + 2NaOH + 3H2O -> 2 Na[Al(OH)4]} \label{eq:2} \tag{2}$$

$$\ce{Al2O3 + 6NaOH + 3H2O -> 2 Na3[Al(OH)6]} \label{eq:3} \tag{3}$$

Look at the above reactions. $\eqref{eq:1}$ and $\eqref{eq:2}$ have different reactants(aluminium and aluminium oxide) but they give the same product.

On the other hand, $\eqref{eq:2}$ and $\eqref{eq:3}$ have same reactants but give different products.

What is exactly going on in these reactions. How do I predict what product is going to be formed in the major amount?

The writers of the book haven't specified the reaction conditions.

Thanks

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  • $\begingroup$ What is strange about different reactants giving the same product or the same reactants giving different products ? It is quite common in chemistry. Chemistry is not mathematics. There are few laws , there are many more or less empirical rules and rest is to be remembered or found if needed. Unless experience teaches you to see unseen behaviour patterns. $\endgroup$
    – Poutnik
    Mar 14 at 11:45
  • $\begingroup$ What is exactly the question? R1 and R2 differ in the productS. R1 involves a redox R2 & R3 don't. Also they're are unbalanced. $\endgroup$
    – Alchimista
    Mar 14 at 11:46
  • $\begingroup$ For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use MathJax $\endgroup$
    – Poutnik
    Mar 14 at 11:48
  • $\begingroup$ Hydrolysis ... is any chemical reaction in which a molecule of water breaks one or more chemical bonds. The term is used broadly for substitution, elimination, and solvation reactions in which water is the nucleophile. Wikipedia $\endgroup$
    – Poutnik
    Mar 14 at 14:51
  • $\begingroup$ Thank you so much Poutnik for formatting and answering to my question. Will use MathJax in future as I have learnt how to use it. $\endgroup$ Mar 15 at 13:23
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I will answer in a more general way to your question, because you have a specific one. Aluminium is absolutely a metal, but it can manifest peculiar features of non-metals. Its oxide, $\text{Al}_2\text{O}_3$ is very inert, while the hydroxide $\text{Al}(\text{OH})_3$ is a light-blue jelly solid with an amphoteric behavior, that is it can react both with acids and bases. I suggest you to read the "bible" Chemistry of the Elements of Greenwood and Earnshaw.

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    $\begingroup$ I was aware of this passive nature of its oxide and amphoteric nature of aluminium metal. However, I was not able to grasp the particular equation. It has now been brought to understanding that it just depends upon the availability of NaOH. In case, the concentrations changes, the products change. Thanks for your advice too. $\endgroup$ Mar 15 at 13:16
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The difference between $(2)$ and $(3)$ is the number of $\ce{NaOH}$ that has been used. If few $\ce{NaOH}$ is available, $\ce{Al2O3}$ reacts according to $(2)$. If much $\ce{NaOH}$ is available, it reacts according to $(3)$. So equation $(3)$ is equal to $(2)$ plus twice the following equation $(4)$ $$\ce{Na[Al(OH)4] + 2 NaOH -> Na3[Al(OH)6]\tag{4}}$$

This is the same for the reaction of metallic aluminum. If only a few $\ce{NaOH}$ is available, the reaction will occur according to $(1)$. If enough $\ce{NaOH}$ is available, it will react according to $(1) + 2·(4)$ giving an equation $(5)$, which is : $$\ce{2 Al + 6 NaOH + 6 H2O -> 2 Na3[Al(OH)6] + 3 H2 \tag{5}}$$

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  • $\begingroup$ So, can I assume that metallic aluminium and aluminium oxide react with NaOH the same way? That's a great explanation BTW. Thanks for explaining. $\endgroup$ Mar 15 at 13:07
  • $\begingroup$ Both $\ce{Al}$ and $\ce{Al2O3}$ react with $\ce{NaOH}$ to produce first $\ce{Na[Al(OH)4]}$ (and of course something else, $\ce{H2}$ or $\ce{H2O}$). Now if enough $\ce{NaOH}$ is available, $\ce{Na[Al(OH)4}$ reacts with it to produce at the end $\ce{Na3[Al(OH)6]}$ $\endgroup$
    – Maurice
    Mar 15 at 15:12

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