# 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?

May 24, 2016

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

$\text{NaOH"(aq) + "HCl"(aq) -> "NaCl"(aq) + "H"_2"O} \left(l\right)$

and

$\text{Ba"("OH")_2(aq) + 2"HCl"(aq) -> "BaCl"_2(aq) + 2"H"_2"O} \left(l\right)$

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 $\text{HCl}$ is $\text{1.19 g/mL}$ 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 $\text{1.00 g/mL}$ as an approximation since the density of dilute aqueous HCl has to be between $1.00$ and $\text{1.49 g/mL}$.
3. Convert both masses to $\text{mol}$s, and compare to see which one is the limiting reagent.
4. Then, convert to $\text{mol}$s of either $\text{NaCl}$ or ${\text{BaCl}}_{2}$ from the $\text{mol}$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 $\text{0.1 g}$ 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 $\text{2.1 mL}$ (its solubility is $\text{4.68 g/100 mL}$ at room temperature), so it shouldn't take that much $\text{HCl}$ to dissolve it.

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

That gives us $3.0 \cancel{\text{g HCl" xx "1 mol"/(36.4609 cancel"g HCl") = "0.08 mols HCl}}$, 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 $\text{mol}$s of sodium hydroxide is approximately ten times the $\text{mol}$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.