What gas is given off when metals react with acid?

Feb 2, 2016

Hydrogen (H2). This was the method used by the first manned-balloon flight using hydrogen (brainchild of Jacques Charles - the namesake for Charles law)

Explanation:

A quarter tonne of sulfuric acid was reacted with half a tonne of iron to fill

https://en.wikipedia.org/wiki/Jacques_Charles#First_hydrogen_balloon

Feb 7, 2016

Hydrogen(H2) gas is given of when metals react with an acid.

Feb 7, 2016

It depends on the metal, but when you work with Aluminum as an example, the reaction becomes:

$\textcolor{b l u e}{\text{2Al} \left(s\right)}$ $\textcolor{b l u e}{+}$ $\textcolor{b l u e}{6 {\text{HCl"(aq) -> 2"AlCl}}_{3} \left(a q\right)}$ $\textcolor{b l u e}{+}$ $\textcolor{b l u e}{3 {\text{H}}_{2} \left(g\right)}$

Another way to see this is the depiction of the reaction mechanism. At a General Chemistry class, you do not need to know this mechanism, but maybe it'll help to get a "behind-the-scenes" on what happens.

What you should notice is that indeed, two $\text{Al}$ atoms begin with react with $\text{HCl}$. Since aluminum has three unpaired electrons, the $\text{Al}$ atoms react by donating singular, radical electrons to cleave the $\text{H"-"Cl}$ bond, and ${\text{H}}_{2}$ gas forms.

After three steps, we form two equivalents of ${\text{AlCl}}_{3} \left(a q\right)$ and a total of three equivalents of ${\text{H}}_{2} \left(g\right)$.

As an interesting aside, if we had two more equivalents of $\text{HCl}$, the reaction would proceed one more step and peak out to be:

$\textcolor{b l u e}{2 \text{Al} \left(s\right)}$ $\textcolor{b l u e}{+}$ $\textcolor{b l u e}{8 {\text{HCl"(aq) -> 2"AlCl}}_{4}^{-} \left(a q\right)}$ $\textcolor{b l u e}{+}$ $\textcolor{b l u e}{3 {\text{H"_2(g) + "2H}}^{+} \left(a q\right)}$

Since aluminum now has no more valence electrons to donate, it must utilize its empty $2 {p}_{z}$ orbital to accept two electrons from ${\text{Cl}}^{-}$, thus leaving behind one equivalent of ${\text{H}}^{+}$ instead of forming ${\text{H}}_{2} \left(g\right)$.