Given the reaction #2"Al"(s) + 6"HCl"(aq) -> 2"AlCl"_3(aq) + 3"H"_2(g)#, if #"0.15 mols"# of #"Al"# is to react with #"0.35 mols"# of #"Cl"#, how many mols of #"Al"# are in excess?

Well, aluminum is the excess reagent, so you have more work to do. I got #"0.03 mols"# of aluminum leftover.

Your reaction is:

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

So from the reaction coefficients, if you have #"2 mols"# of #"Al"#, you need #"6 mols"# of #"HCl"# to react exactly without leftover reactants.

Moles are extensive, so:

• if you have #"1 mol"# of #"Al"#, you would need #"3 mols"# of #"HCl"#.
• if you have #"0.5 mols"# of #"Al"#, you would need #"1.5 mols"# of #"HCl"#.
• if you have #"0.15 mols"# of #"Al"#, you would need #"0.45 mols"# of #"HCl"#.

Notice how each time, the mols of #"Al"# are one-third the mols of #"HCl"#. You actually only have #"0.35 mols"# of #"HCl"#, so #"HCl"# is your limiting reagent. You have #"0.10 mols"# less of it than you would need.

Since that is the case, and the mols of #"Al"# need to be one-third of the mols of #"HCl"#, if you have #"0.35 mols"# of #"HCl"#, you would need #"0.12 mols"# or so of #"Al"# to react exactly:

#0.35 cancel"mols HCl" xx "2 mols Al"/(6 cancel"mols HCl") = "0.1167 mols Al" ~~ "0.12 mols Al"#

So to two decimal places, your remaining #"Al"# excess reagent is:

#"0.15 mols Al total"# #-# #"0.1167 mols Al consumed"#

#=# #ulcolor(blue)bb"0.03 mols Al leftover"#