If at a certain temperature, #"6.8 g"# of #"NH"_3# completely decomposes to form #"H"_2(g)# and #"N"_2(g)#, how many mols of #"Al"(s)# would be needed to react with excess aqueous #"HCl"# to form the same amount of #"H"_2(g)#?

The idea is to see how many mols of #"H"_2# you would have made in one reaction and reuse that value in the context of the other reaction.

Assuming you mean #"6.8 g"# of #"NH"_3# decomposes as:

#2"NH"_3(g) rightleftharpoons 3"H"_2(g) + "N"_2(g)#

and assuming the reaction goes to completion, we make:

#"6.8 g NH"_3 xx ("1 mol NH"_3)/(14.007 + 3xx1.0079 "g NH"_3) xx ("3 mols H"_2)/("2 mol NH"_3)#

#= "0.5989 mols H"_2#

To make that many mols of #"H"_2# in the reaction of

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

You'll need

#"0.5989 mols H"_2(g) xx ("1 mol Al"(s))/("3 mols H"_2(g))#

#=# #"0.1996 mols Al"#

To 2 sig figs, it would be #color(blue)("0.20 mols Al")#.

The forward ammonia reaction was called the Haber process, i.e. the production of #"NH"_3(g)# from #"N"_2(g)# and #"H"_2(g)# at high pressures and temperatures. We looked at the reverse reaction.

The aluminum reaction is a common example of placing a metal into water and reacting it with a strong acid to produce a gas. Since your strong acid was #"HCl"#, you made #"H"_2# gas.

(If you had used #"HNO"_3#, you might have made the brown #"NO"_2# gas.)