What is/are a practice problem(s) that involves the following topics? Concentration, limiting/excess reactants, acids/bases, solubility

Well, sure, let's hit all of the above using a single idea: a strong acid like #"H"_2"SO"_4# can oxidize a metal below it in the standard reduction potential table.

One option (of many) is #"Fe"# being oxidized to #"Fe"^(2+)#. In effect, we end up dissolving it. So, consider the following reaction:

#"H"_2"SO"_4(aq) + "Fe"(s) -> "Fe"^(2+)(aq) + "SO"_4^(2-)(aq) + "H"_2(g)#

1. Suppose the concentration of the sulfuric acid is #"2.0 M"# (in water) and you have #"3.50 g"# of #"Fe"(s)#. By using #"27.00 mL"# of the acid, the iron dissolves. Determine which substance is the limiting reactant.

2. Calculate the mols of substance in excess, and note which one it was. Then, determine the mols of #"Fe"^(2+)# in solution.

3. Are any products Lewis acids or bases? (Hint: #"Fe"^(2+)# forms #["Fe"("OH"_2)_6]^(2+)# in aqueous solution.)

4. The excess #"Fe"(s)# is removed with tweezers. The solution is transferred to a larger container, and water is added until the solution reaches #"500.00 mL"#.

If #"3.205 g"# of #"NaOH"# (#"FW"# #=# #"39.9959 g/mol"#) is then added to the solution, one should saturate the solution with #"Fe"("OH")_2(s)#, but also have some remaining #"Fe"^(2+)# in solution.

Calculate the resultant solubility of #"Fe"("OH")_2(s)# in the aqueous solution (ignoring the excess #"Fe"^(2+)# remaining) at #20^@ "C"# in #"g/100 mL soln"#.

And that checks off all the points: concentration, limiting reactants/excess, acids/bases, solubility.

#1)#

Try getting both reactants into mols.

#("2.0 mols H"_2"SO"_4)/cancel"L" xx cancel"1 L"/(1000 cancel"mL") xx 27.00 cancel"mL" = "0.0540 mols H"_2"SO"_4#

#3.50 cancel"g Fe" xx "1 mol"/(55.845 cancel"g Fe") = "0.0627 mols Fe"#

Since #"Fe"# is #1:1# with #"H"_2"SO"_4#, the sulfuric acid is the limiting reactant.

#2)#

The iron is thus in excess by

#"0.0627 mols" - "0.0540 mols" = color(blue)("0.0087 mols")#, or #"0.484 g"#.

From the stoichiometry of the reaction,

#"0.0540 mols H"_2"SO"_4 = color(blue)("0.0540 mols Fe"^(2+))#

#3)#

Yes, the iron(II) cation is a Lewis acid, because it has an empty #4s# orbital that can accept an electron pair, specifically from the water molecules in aqueous solution (which act as Lewis bases).

#4)#

#"Fe"^(2+)(aq) + 2"OH"^(-)(aq) -> "Fe"("OH")_2(s)#

#3.205 cancel"g NaOH" xx "1 mol"/(39.9959 cancel"g NaOH") = "0.0801 mols NaOH"#,

which reacts with #"0.0401 mols"# of #"Fe"^(2+)# to form

#0.0401 cancel("mols Fe"("OH")_2(s)) xx ("89.86 g Fe"("OH")_2(s))/cancel"1 mol"#

#=# #"3.60 g Fe"("OH")_2(s)# per #"500.00 mL"# of aqueous solution,

or #color(blue)("0.72 g/100 mL")# solution.

(#"0.0139 mols"# of #"Fe"^(2+)# was in excess, which then has a #"0.0278 M"# concentration in solution.)