Unveiling the Water-to-Salt Ratio in Barium Chloride Hydrate

Step 1 of 2

The goal of the problem is to calculate the value of \(x\) in \(\mathrm{BaCl}_{2} \cdot x \mathrm{H}_{2} \mathrm{O}\)

Given:

\(\begin{array}{l}\text{Mass of the compound } =1.936 \mathrm{~g}\\ \text{Mass of anhydrous } \mathrm{BaSO}_{4}=1.864 \mathrm{~g}\end{array}\)

Here, the compound is treated with sulfuric acid, and \(\mathrm{BaCl}_{2}\) dissolves losing its waters of hydration. The balanced equation for the reaction occurring is shown below:

\(\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{BaSO}_{4}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq})\)

The mass of the water formed is the difference between the mass of the hydrated compound and the mass of anhydrous \(\mathrm{BaCl}_{2}\).

First, let’s calculate the mass of anhydrous \(\mathrm{BaCl}_{2}\) based on the amount of \(\mathrm{BaSO}_{4}\) produced:

\(\begin{array}{l}\text{Molar mass of } \mathrm{BaSO}_{4}=233.38 \mathrm{~g} / \mathrm{mol}\\ \text{Molar mass of } \mathrm{BaCl}_{2}=208.23 \mathrm{~g} / \mathrm{mol}\end{array}\)

\(\begin{array}{l} =1.864 \mathrm{~g} \mathrm{BaSO}_{4} \times \frac{1 \mathrm{~mol} \mathrm{BaSO}_{4}}{233.38 \mathrm{~g} \mathrm{BaSO}_{4}} \times \frac{1 \mathrm{~mol} \mathrm{BaCl}_{2}}{1 \mathrm{~mol} \mathrm{BaSO}_{4}} \times \frac{208.23 \mathrm{~g} \mathrm{BaCl}_{2}}{1 \mathrm{~mol} \mathrm{BaCl}_{2}} \\ =1.663 \mathrm{~g} \mathrm{BaCl}_{2} \end{array}\)

Now that we have the mass of both hydrated and anhydrous compound, let’s calculate the mass of the water:

\(\begin{array}{l} =1.936 \mathrm{~g}-1.663 \mathrm{~g} \\ =0.273 \mathrm{~g} \text { of } \mathrm{H}_{2} \mathrm{O} \end{array}\)