If an unknown solid is either #"NaOH"# or #"Ba"("OH")_2#, and it is reacted with excess #37%"w/w"# stock #"HCl"# (#rho = "1.19 g/mL"#), can you still determine which base is the limiting reagent?

I believe it's a valid test. The reactions would be

#"NaOH"(aq) + "HCl"(aq) -> "NaCl"(aq) + "H"_2"O"(l)#

and

#"Ba"("OH")_2(aq) + 2"HCl"(aq) -> "BaCl"_2(aq) + 2"H"_2"O"(l)#

Unfortunately, none of the products are sufficiently insoluble to form a precipitate in a laboratory setting, as far as I can tell.

You can indeed find a limiting reagent for this.

1. Start from the mass of the base, and suppose you have one or the other.
2. Then, note the density of #37%# w/w stock #"HCl"# is #"1.19 g/mL"# and solve for its mass in grams. You only need an estimate, I suppose, so if your HCl is dilute, it's OK to assume the density is #"1.00 g/mL"# as an approximation since the density of dilute aqueous HCl has to be between #1.00# and #"1.49 g/mL"#.
3. Convert both masses to #"mol"#s, and compare to see which one is the limiting reagent.
4. Then, convert to #"mol"#s of either #"NaCl"# or #"BaCl"_2# from the #"mol"#s of limiting reagent.

If you have the experimental mass of the product, then you should be able to see two clearly different results for the theoretical mass from solving either reaction 1 or reaction 2.

Suppose you had #"0.1 g"# of base. Then you have:

#"0.1" cancel("g NaOH") xx "1 mol"/(39.9959 cancel("g")) = color(blue)(2.5xx10^(-3) "mols NaOH")#

#"0.1" cancel("g Ba"("OH")_2) xx "1 mol"/(171.3138 cancel("g")) = color(blue)(5.8xx10^(-4) "mols Ba"("OH")_2)#

If you had barium hydroxide at room temperature, a reasonable amount of water to dissolve it is #"2.1 mL"# (its solubility is #"4.68 g/100 mL"# at room temperature), so it shouldn't take that much #"HCl"# to dissolve it.

So let's just say we had #"3.0 mL"# dilute #"HCl"#. Then the mass is about #3.0 cancel"mL" xx (~"1.00 g")/cancel("mL") = "3.0 g"#.

That gives us #3.0 cancel"g HCl" xx "1 mol"/(36.4609 cancel"g HCl") = "0.08 mols HCl"#, which would naturally make it the excess reactant.

Thus, the base can easily be the limiting reagent.

Since for the same mass of base, the #"mol"#s of sodium hydroxide is approximately ten times the #"mol"#s of barium hydroxide, simply calculate the mass of the product both ways, and whichever mass is closer should be correct, if the masses you wrote down were correct.

A factor of #10# is quite significant.