1
$\begingroup$

Get a metal like sodium and then you burn it to make sodium oxide because burning something is adding oxygen. Then you add the sodium oxide to an acid like hydrochloric acid. That should make a salt (sodium chloride) and water. Evaporate or filter the water away and you are just left with the salt itself.

Is this possible to do or am I just completely wrong?

$\endgroup$
2
  • $\begingroup$ Why do you even think it could be wrong? $\endgroup$
    – Mithoron
    May 16, 2016 at 19:43
  • 1
    $\begingroup$ You can't filter anything out of a solution. Other than that, it's a feasible plan. $\endgroup$ Jun 3, 2016 at 22:12

3 Answers 3

3
$\begingroup$

Your description is almost correct - note that when sodium metal burns in air, it forms primarily sodium peroxide in addition to sodium oxide. Sodium peroxide will react with hydrochloric acid to give sodium chloride and hydrogen peroxide. Heating will decompose that to water plus oxygen. You should still be able to get the sodium chloride product that you expected.

$\endgroup$
2
$\begingroup$

Fizzle Nizzle's idea is nearly correct. First, all metals can be oxidized and form oxides. Second, all oxides can react with acids like $\ce{HCl}$. And a salt (like a metallic chloride) is always obtained in solution. But ! But it is not always possible to save the obtained chloride out of this solution. By evaporating the solution, the salt is often hydrolyzed, and by heating it, its solution sometimes may undergo the reverse reaction of its formation. This is particularly true for metals oxidized to the oxidation number +III or higher.

Aluminium is a good example of this behavior. Aluminium can be oxidized into aluminium oxide $\ce{Al2O3}$ (and into hydroxide $\ce{Al(OH)3}$ if water is present). These oxides or hydroxides can react with $\ce{HCl}$ solutions to get a solution of $\ce{Al(III)}$ ions, the formula of which can be written as $\ce{Al^{3+}}$ or $\ce{[Al^{3+}(H2O)_n]}$. But its evaporation will not produce $\ce{AlCl3·nH2O}$, whatever the value of $n$. Its evaporation will have the ion $\ce{Al^{3+}}$ get hydrolyzed (= react with water) and produce successfully $\ce{AlCl2(OH)}$, $\ce{AlCl(OH)2}$, then $\ce{Al(OH)3}$ each time with more $\ce{HCl}$ relieved in the vapor, in equations like : $$\ce{Al^{3+}+ +3 Cl^- + H2O -> AlCl2(OH)(s) + HCl(g)}$$ $$\ce{AlCl2OH + H2O -> AlCl(OH)2(s) + HCl(g)}$$ $$\ce{AlCl(OH)2 + H2O -> Al(OH)3(s) + HClg)}$$ If the non hydrolyzed compound $\ce{AlCl3}$ is to be obtained, this must be done by another way, and not by neutralization of an hydroxide by an acid. It may be done by direct syntheses, and for example by the reaction $$\ce{2Al + 3 Cl2 -> 2 AlCl3}$$ or by indirect ways avoiding the presence of water like : $$\ce{Al2O3 + 3 C + 3 Cl2 -> 2 AlCl3 + 3 CO}$$

$\endgroup$
0
$\begingroup$

You are correct. Metal oxides will react with acids to form water and the corresponding salt. It's the same with metal hydroxides, they also react with an acid to form water and the salt.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.