What gas is given off when metals react with acid?

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

#color(blue)("2Al"(s))# #color(blue)(+)# #color(blue)(6"HCl"(aq) -> 2"AlCl"_3(aq))# #color(blue)(+)# #color(blue)(3"H"_2(g))#

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 #"Al"# atoms begin with react with #"HCl"#. Since aluminum has three unpaired electrons, the #"Al"# atoms react by donating singular, radical electrons to cleave the #"H"-"Cl"# bond, and #"H"_2# gas forms.

After three steps, we form two equivalents of #"AlCl"_3(aq)# and a total of three equivalents of #"H"_2(g)#.

As an interesting aside, if we had two more equivalents of #"HCl"#, the reaction would proceed one more step and peak out to be:

#color(blue)(2"Al"(s))# #color(blue)(+)# #color(blue)(8"HCl"(aq) -> 2"AlCl"_4^(-)(aq))# #color(blue)(+)# #color(blue)(3"H"_2(g) + "2H"^(+)(aq))#

Since aluminum now has no more valence electrons to donate, it must utilize its empty #2p_z# orbital to accept two electrons from #"Cl"^(-)#, thus leaving behind one equivalent of #"H"^(+)# instead of forming #"H"_2(g)#.