Question

After the second heating a sample of BaSO, had a mass of 2.22 g. If the filter paper weighed 1.58 grams what is the weight of the dry BaSO? Show work. a. Molar mass of Baso 233-4 almol b. Molar mass of BaCl2_ 208.23 al mol c. Moles of BaSO4 collected 233-4 d. Moles of BaCl2 in the original solution e. Grams of BaCl, in the original solution

Answer 1

The experiment is suggesting that it is precipitation of BaSO₄ from BaCl₂ using any SO₄^2- ion reagent. Whatever be it, we know from one mol of BaCl₂ we can get only 1 mol of BaSO₄.

Now mass of BaSO₄ collected = mass of BaSO₄ with filter paper - mass of filter paper = 2.22 g - 1.58 g = 0.64 g.

Thus, moles of BaSO₄ = (mass of BaSO₄ / molar mass of BaSO₄ ) = (0.64 / 233.4) mol = 2.742 x 10^-3 mol (c).

Now, moles of BaCl₂ in the original solution = moles of BaSO₄ precipitated = 2.742 x 10^-3 mol (d).

Grams of BaCl₂ in original solution = moles of BaCl₂ x molar mass of BaCl₂ = (2.742 x 10^-3 x 208.23) g = 0.571 g (e).